· Exercise 1. 3 -3 Exercise 1. 3 -3 er r s er e e si s 3 -i e ice e si s - 1 1 16. stru t 1- 1 Exercise 1.1 1-1 Exercise 1. 1-1 Exercise 1.3 1 -1 1. $er"s rmu -

CONTENTS

1. Number System 1-16

Exercise 1.1 1-3

Exercise 1.2 3-5

Exercise 1.3 5-6

Exercise 1.4 6-10

Exercise 1.5 10-13

Exercise 1.6 13

Multiple Choice Questions 13-16

2. Exponents of Real Numbers 17-48

Exercise 2.1 17-24

Exercise 2.2 24-38

Very Short Answer Type Questions 38-39


3. Rationalisation 49-66

Exercise 3.1 49-50

Exercise 3.2 50-59



4. Algebraic Identities 67-91

Exercise 4.1 67-70

Exercise 4.2 71-73

Exercise 4.3 73-79

Exercise 4.4 79-83

Exercise 4.5 83-84



5. Factorization of Algebraic Expressions 92-105

Exercise 5.1 92-96

Exercise 5.2 96-98

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Exercise 5.3 98

Exercise 5.4 99-101



6. Factorization of Polynomials 106-133

Exercise 6.1 108-109





Very Short Answer Type Questions 129


7. Linear Equations in Two Variables 134-153

Exercise 7.1 134






8. Co-ordinate Geometry 154-157



9. Introduction to Euclid’s Geometry 158-164



10. Lines and Angles 165-204

Exercise 10.1 173-175

Exercise 10.2 175-181

Exercise 10.3 181-185

Exercise 10.4 185-194

Very Short Answer Type Questions 194



11. Triangle and its Angles 205-234

Exercise 11.1 210-213

Exercise 11.2 213-220



12. Congruent Triangles 235-269

Exercise 12.1 245-249

Exercise 12.2 249-250

Exercise 12.3 250-254

Exercise 12.4 254

Exercise 12.5 254-257

Exercise 12.6 257-261



13. Quadrilaterals 270-309

Exercise 13.1 281-282

Exercise 13.2 282-285

Exercise 13.3 285-288

Exercise 13.4 288-297



14. Areas of Parallelograms and Triangles 310-347

Exercise 14.1 316-317

Exercise 14.2 317-318

Exercise 14.3 319-337



15. Circles 348-411

Exercise 15.1 373

Exercise 15.2 373-379

Exercise 15.3 379-380


Exercise 15.4 380-387

Exercise 15.5 387-398



16. Construction 412-421

Exercise 16.1 412-414

Exercise 16.2 414-418

Exercise 16.3 418-421

17. Heron’s Formula 422-440

Exercise 17.1 423-427

Exercise 17.2 427-433



18. Surface Area and Volume of a Cuboid and Cube 441-459

Exercise 18.1 441-446

Exercise 18.2 446-453



19. Surface Area and Volume of A Right Circular Cylinder 460-478

Exercise 19.1 460-464

Exercise 19.2 464-473



20. Surface Area and Volume of A Right Circular Cone 479-496

Exercise 20.1 479-484

Exercise 20.2 485-490



21. Surface Area and Volume of A Sphere 497-517

Exercise 21.1 497-501

Exercise 21.2 501-511




22. Tabular Representation of Statistical Data 518-532

Exercise 22.1 520-527

Exercise 22.2 527-531


23. Graphical Representation of Statistical Data 533-569

Exercise 23.1 535-549

Exercise 23.2 549-563

Exercise 23.3 563-568

Multiple Choice Questions 569

24. Measures of Central Tendency 570-598

Exercise 24.1 575-582

Exercise 24.2 582-589

Exercise 24.3 589-592

Exercise 24.4 592-593



25. Probability 599-610

Exercise 25.1 599-607




1 Arundeep’s Mathematics (R.D.) 9th

NUMBER SYSTEM

Points to Remember :

1. Natural number : Numbers like 1, 2, 3, 4,

5, ..... are called natural numbers. There

are also called counting numbers. The set

of natural number is denoted by N.

2. Whole numbers : Numbers like 0, 1, 2, 3,

4, ..... are called whole numbers. The set

of whole numbers is denoted by W.

3. Integers : The negative integers together

whole numbers are called integers : ....., –

4, –3, –2, –1, 0, 1, 2, 3, 4, ..... are integers.

The set of integers is denoted by Z.

4. Rational numbers : The number in the

form of q

p where p and q are integers and

q 0, is called rational numbers like 5

2,

7

1, 0,

4

3,

8

7,

4

9 etc.

The set of rational numbers is denoted by

Q.

Q = q

p : p and q z and q 0.

5. Irrational numbers : The numbers which

are not rationals, are called irrational

numbers. e.g. 2 , 3 , 5 etc.

6. Real numbers : Rational numbers and

irrational numbers together are called real

numbers.

7. Number line and representation of

numbers on a number line : A line on

which numbers are shown, is called a

number line.

We can represent real numbers on the

number line.

8. Decimal expansion of rational numbers:

The decimal of rational numbers are either

terminating or non-terminating recurring

decimals and decimal of irrational numbers

are non-terminating non-recurring.

9. If r is a rational number and s is irrational

then r + s, r – s and s

r are irrational

numbers.

10. Finding rational numbers between two

numbers : We can find an infinite number

of rational numbers between two numbers.

Let a and b are two numbers, then a rational

number between then = 2

ba .

11. Some useful results on irrational

numbers :

(i) Negative of an irrational number is an

irrational number.

(ii) The sum of a rational number and an

irrational number is an irrational number.

(iii) The product of a non-zero rational number

and an irrational number is an irrational

number.

(iv) The sum, difference, product and quotient

of two irrational numbers need not be an

irrational number.

EXERCISE 1.1

1. Is zero a rational number? Can you write it

in the form q

p, where p and q are integers

and q 0? [NCERT]

Sol. Yes, zero is a rational number e.g.

11

0,

12

0,

31

0, etc.

2. Find five rational numbers between 1 and

2. [NCERT]

Sol. We know that one rational number between

two numbers a and b = 2

ba

1



Therefore one rational number between 1

and 2

= 2

21 =

2

3

Similarly, rational number between 1 and

2

3

= 2

1

2

3

1

1 =

2

1

2

32 =

2

1 ×

2

5 =

4

5

Rational number between 2

3 and 2

= 2

1

1

2

2

3 =

2

1

2

43

= 2

1 ×

2

7 =

4

7

Three rational number between 1 and 2

are 4

5,

2

3,

4

7 or

4

5,

4

6,

4

7

and rational number between 1 and 4

5

= 2

1

4

51

= 2

1

4

54 =

2

1 ×

4

9 =

8

9

and rational number between 4

7 and 2

= 2

1

1

2

4

7 =

2

1

4

87

= 2

1 ×

4

15 =

8

15

Hence five rational numbers are 8

9,

4

5,

2

3,

4

7,

8

15

3. Find six rational numbers between 3 and 4.

[NCERT]

Sol. One rational number between 3 and 4

= 2

1(3 + 4) =

2

7

One rational number between 3 and 2

7

= 2

1

2

73 =

2

1

2

76

= 2

1 ×

2

13 =

4

13

One rational number between 2

7 and 4

= 2

1

1

4

2

7 =

2

1

2

87

= 2

1

2

15 =

4

15


13

= 2

1

4

133 =

2

1

4

1312

= 2

1 ×

4

25 =

8

25

One rational number between 4

15 and 4

= 2

1

1

4

4

15 =

2

1

4

1615

= 2

1 ×

4

31 =

8

31


25



= 2

1

8

253

= 2

1

8

2524 =

2

1 ×

8

49 =

16

49

Six rational numbers between 3 and 4 are

16

49,

8

25,

4

13,

2

7,

4

15,

8

31

4. Find five rational numbers between 5

3 and

5

4.

Sol. _ Rational number between a and b

= 2

1(a + b)


3 and

5

4

= 2

1

5

4

5

3 =

2

1 ×

5

7 =

10

7


3 and

10

7

= 2

1

10

7

5

3 =

2

1

10

76

= 2

1 ×

10

13 =

20

13


7 and

5

4

= 2

1

5

4

10

7 =

2

1

10

87

= 2

1 ×

10

15 =

20

15 =

4

3


3 and

20

13

= 2

1

20

13

5

3 =

2

1

20

1312

= 2

1 ×

20

25 =

40

25 =

8

5

and rational number between 4

3 and

5

4

= 2

1

5

4

4

3 =

2

1

20

1615

= 2

1 ×

20

31 =

40

31

Required rational numbers are

8

5,

20

13,

10

7,

4

3,

40

31

5. Are the following statements true or false?

Give reason for your answer.

(i) Every whole number is a natural number.

[NCERT]

(ii) Every integer is a rational number.

(iii) Every rational number is an integer.

(iv) Every natural number is a whole number.

(v) Every integer is a whole number.

(vi) Every rational number is a whole number.

Sol. (i) False, as 0 is not a natural number.

(ii) True.

(iii) False, as 2

1,

3

1 etc. are not integers.

(iv) True.

(v) False, _ negative natural numbers are not

whole numbers.

(vi) False, _ proper fraction are not whole

numbers.

EXERCISE 1.2

1. Express the following rational numbers as

decimals :

(i)100

42(ii)

500

327



(iii)4

15

Sol. (i) 100

42 = 0.42

(ii)500

327 =

2500

2327

=

1000

654 = 0.654

(iii)4

15 =

254

2515

=

100

375 = 3.75

2. (i) 3

2(ii) –

9

4

(iii)15

2(iv) –

13

22

(v)999

437(vi)

26

33

Sol. (i) 3

2 = 0.66..... = 6.0

0.66.....3)2.000( 18 20 18 20 18 2

(ii) –9

4 = –(0.444.....) = 4.0

0.444.....9)4.000( 36 40 36 40 36 4

(iii)15

2 = –(0.133.....) = 31.0

0.133.....15)2.000( 15 50 45 50 45 5

(iv) –13

22 = –1

13

9

= –1.692307692307..... = 692307.1

0.692307.692307.....13)9.000000000000( 78 120 117 30 26 40 39 100 91 90 78 120 117 30 26 40 39 1

(v)999

437 = 0.437437..... = 437.0

0.437437.....999)437000000( 3996 3740 2997 7430 6993 4370 3996 3740 2997 7430 6993 437



(vi)26

33 = 1

26

7 = 1.2692307692307.....

= 6923072.1

0.2692307692307.....26)7.000000000000( 52 180 156 240 234 60 52 80 78 200 182 180 156 240 234 60 52 80 78 200 182 18

3. Look at several examples of rational

numbers in the form q

p(q 0), where p

and q are integers with no common factors

other than 1 and having terminating decimal

representations. Can you guess what

property q must satisfy?

Sol. The property is if the denominators have

factors 2 or 5 or both, the decimal

representation will be terminating e.g.

2

1 = 0.5,

4

1 = 0.25,

5

1 = 0.2,

10

1 = 0.1,

20

1 = 0.5,

8

3 = 0.125,

40

3 = 0.075 etc.

EXERCISE 1.3

1. Express each of the following decimals in

the form q

p:

(i) 0.39 (ii) 0.750

(iii) 2.15 (iv) 7.010

(v) 9.90 (vi) 1.0001

Sol. (i) 0.39 = 100

39

(ii) 0.750 = 1000

750 =

4

3

(iii) 2.15 = 100

215 =

20

43

(iv) 7.010 = 71000

10 = 7

100

1 =

100

701

(v) 9.90 = 100

990 =

10

99

(vi) 1.0001 = 10000

10001

2. Express each of the following decimals in

the form q

p:

(i) 4.0 (ii) 37.0

(iii) 54.0 (iv) 621.0

(v) 3.125 (vi) 7.4

(vii) 74.0 [NCERT]

Sol. (i) 4.0

Let x = 4.0 = 0.4444... ...(i)

10x = 4.4444... ...(ii)



Subtracting (i) from (ii)

9x = 4 x = 9

4

4.0 = 9

4

(ii) 37.0

Let x = 37.0 = 0.373737... ...(i)

100x = 37.373737... ...(ii)


99x = 37 x = 99

37

37.0 = 99

37

(iii) 54.0

Let x = 54.0 = 0.545454... ...(i)

100x = 54.545454... ...(ii)


99x = 54 x = 99

54

x = 11

6

54.0 = 11

6

(iv) 621.0

Let x = 621.0 = 0.621621621... ...(i)

1000x = 621.621621621... ...(ii)


999x = 621 x = 999

621

x = 37

23

(Dividing the numerator and denominator

by 27)

(v) 3.125

Let x = 3.125 = 125.333... ...(i)

10x = 1253.333... ...(ii)


9x = 1253 – 125 = 1128

x = 9

1128 =

3

376

3.125 = 3

376

(vi) 7.4

Let x = 7.4 = 4.777... ...(i)

10x = 47.777... ...(ii)


9x = 47 – 4 = 43

x = 9

43

7.4 = 9

43

(vii) 74.0

Let x = 74.0

10x = 7.4 = 4.777... ...(i)

and 100x = 47.777... ...(ii)


90x = 43 x = 90

43

74.0 = 90

43

EXERCISE 1.4

1. Define an irrational number.

Sol. A number which cannot be expressed in

the form of q

p where p and q are integers

and q 0 is called an irrational number.



2. Explain, how irrational numbers differ from

rational numbers?

Sol. A rational number can be expressed in either

terminating decimal or non-terminating

recurring decimals but an irrational number

is expressed in non-terminating non-

recurring decimals.

3. Examine, whether the following numbers

are rational or irrational:

(i) 7 (ii) 4

(iii) 2 + 3 (iv) 3 + 2

(v) 3 + 5 (vi) 2)22(

(vii) )22( )22( (viii) 2)32(

(ix) 5 – 2 (x) 23 [NCERT]

(xi) 225 [NCERT] (xii) 0.3796 [NCERT]

(xiii) 7.478478..... [NCERT]

(xiv) 1.101001000100001..... [NCERT]

Sol. (i) 7

It is an irrational as 7 is not a perfect square.

(ii) 4

It is a rational number as 4 is a perfect

square of 2.

(iii) 2 + 3

It is an irrational number as sum of a rational

number and an irrational number is also an

irrational number.

(iv) 3 + 2

Irrational as sum of two irrational numbers

is also an irrational number.

(v) 3 + 5 is an irrational number as sum of

two irrational numbers is also an irrational.

(vi) 2)22( = 2 + 4 + 2 2 × 2 = 6 + 4 2

{_ (a – b)2 = a2 + b2 – 2ab)}

It is an irrational number as sum of a rational

and an irrational number is an irrational

number.

(vii) )22( )22( = (2)2 – ( 2 )2

{_ (a + b) (a – b) = a2 – b2}

= 4 – 2 = 2

Which a rational number

(viii) 2)32( = 2 + 3 + 2 2 3

{_ (a + b)2 = a2 + b2 + 2ab}

= 5 + 2 6

Which is an irrational number as sum of a

rational and an irrational number is an

irrational number.

(ix) 5 – 2 is an irrational number as difference

of an irrational number and a rational

number is also an irrational number.

(x) 23 is an irrational number as 23 is not

a perfect square.

(xi) 225 = 15

Which is a rational number.

(xii) 0.3796 is a rational number its decimal is

terminating.

(xiii) 7.478478..... = 478.7

Which is non-terminating recurring decimal.

Therefore it is a rational number.

(xiv) 1.101001000100001.....

It is an irrational number as its decimal is

non-terminating non-recurring decimal.

4. Identify the following as rational or irrational

numbers. Give the decimal representation

of rational numbers:

(i) 4 (ii) 3 18

(iii) 44.1 (iv)27

9

(v) – 64 (vi) 100

Sol. (i) 4 = 2

It is a rational number.



(ii) 3 18 = 3 × 29 = 3 × 3 2 = 9 2

It is an irrational number.

(iii) 44.1 = 2.12.1 = 1.2


(iv)27

9 =

3

1 =

3

1

= 33

31

=

3

3

It is an irrational number.

(v) – 64 = – 88 = –8


(vi) 100 = 1010 = 10


5. In the following equations, find which

variables x, y, z etc. represent rational or

irrational numbers:

(i) x2 = 5 (ii) y2 = 9

(iii) z2 = 0.04 (iv) u2 = 4

17

(v) v2 = 3 (vi) w2 = 27

(vii) t2 = 0.4

Sol. (i) x2 = 5 x2 = (+ 5 )2

x = 5

Which is an irrational number.

(ii) y2 = 9 y2 = (3)2

y = 3


(iii) z2 = 0.04 = (0.2)2

z = 0.2


(iv) u2 = 4

17 =

2

2

17

u = 2

17


(v) v2 = 3 = ( 3 )2

v2 = 3


(vi) w2 = 27 = 9 × 3 = (3 3 )2

w = 3 3


(vii) t2 = 0.4 t2 = 10

4 =

2

10

2

t = 10

2 =

1010

102

= 10

102 =

5

10


6. Given two rational numbers lying between

0.232332333233332... and

0.212112111211112.

Sol. Two rational numbers lying between

0.232332333233332... and

0.212112111211112...

will be 0.232 and 0.212

7. Give two rational numbers lying between

0.515115111511115... and 0.5353353335...

Sol. Two rational numbers lying between

0.515115111511115... and

0.535335333533335...

will be 0.515, 0.535

8. Find one irrational numbers between 0.2101

and 0.2222... = 2.0 .

Sol. One irrational number lying between 0.2101

and 0.2222... = 2.0 will be 2201.0010001...

9. Find a rational number and also an irrational

number lying between the numbers,

0.3030030003... and 0.3010010001...

Sol. Between two numbers 0.3030030003... and

0.3010010001..., a rational will be 0.301 and



irrational number will be 0.3020020002...

10. Find three different irrational numbers

between the rational numbers 7

5 and

11

9.

[NCERT]

Sol. Let a = 7

5 = 0.714285714285

= 714285.0

and b = 11

9 = 0.8181... = 81.0

0.714285717)5.00000000( 49 10 7 30 28 20 14 60 56 40 35 50 49 10 7 3

0.818111)9.0000( 88 20 11 90 88 20 11 9

We see that a has six digits and b has two

digits, i.e. 0.71 < 0.81

7

5 <

11

9

First irrational number

will be 0.72020020002...

and 0.74020020002...

and 0.76020020002...

11. Give an example of each, of two irrational

numbers whose:

(i) difference is a rational number.

(ii) difference is an irrational number.

(iii) sum is a rational number.

(iv) sum is an irrational number.

(v) product is a rational number.

(vi) product is an irrational number.

(vii) quotient is a rational number.

(viii)quotient is an irrational number.

Sol. (i) Two numbers whose difference is also

a rational number. e.g. 2 , 2 which are

irrational numbers.

Difference = 2 – 2 = 0 which is also a

rational number.

(ii) Two numbers whose difference is an

irrational number.

e.g. 3 and 2 which are irrational

numbers.

Now difference = 3 – 2 which is also

an irrational number.

(iii) Let two irrational numbers be 3 and

– 3 which are irrational numbers.

Now sum = 3 + (– 3 ) = 3 – 3 = 0


(iv) Let two numbers be 5 , 3 which are

irrational numbers.

Now sum = 5 + 3 which is an irrational

number.

(v) Let numbers be 3 + 2 and 3 – 2

which are irrational numbers.

Now product = ( 3 + 2 ) ( 3 – 2 )

= 3 – 2 = 1 which is a rational number.

(vi) Let numbers be 3 and 5 , which are

irrational number.

Now product = 3 × 5 = 53 = 15

which is an irrational number.



(vii) Let numbers be 6 2 and 2 2 which are

irrational numbers.

Quotient = 22

26 = 3 which is a rational

number.

(viii)Let numbers be 3 and 5 which are

irrational numbers.

Now quotient = 5

3 =

5

3 which is an

irrational number.

12. Find two irrational numbers between 0.5

and 0.55.

Sol. Two irrational numbers between 0.5 and

0.55 will be 0.51010010001... and

0.52020020002...

13. Find two irrational numbers lying between

0.1 and 0.12.

Sol. Two irrational numbers lying between 0.1

and 0.12 will be 0.1010010001... and

0.1020020002...

14. Prove that 3 + 5 is an irrational number..

Sol. Let 3 + 5 be a rational number,,

and Let x = 3 + 5

Squaring both sides,

x2 = ( 3 + 5 )2 = 3 + 5 + 2 × 3 × 5

x2 = 8 + 2 15

x2 – 8 = 2 15

2

82 x

= 15

_ x is rational

2

82 x

will be rational

15 will be rational

But it is not possible as 15 is an irrational

number.

Our supposition is wrong

Hence 3 + 5 is an irrational number..

EXERCISE 1.5

1. Complete the following sentences:

(i) Every point on the number line corresponds

to a ... number which many be either ... or

...

(ii) The decimal form of an irrational number

is neither ... nor ...

(iii) The decimal representation of a rational

number is either ... or ...

(iv) Every real number is either ... number or

... number.

Sol. (i) Every point on the number line

corresponds to a real number which many

be either rational or irrational.

(ii) The decimal form of an irrational number

is neither terminating nor repeating.

(iii) The decimal representation of a rational

number is either terminating or non-

terminating, recurring.

(iv) Every real number is either rational number

or an irrational number.

2. Find whether the following statements are

true or false:

(i) Every real number is either rational or

irrational .

(ii) is an irrational number.

(iii) Irrational numbers cannot be represented

by points on the number line.

Sol. (i) True.

(ii) True. (Value of = 3.14)

(iii) False : we can represent irrational number

also.

3. Represent 6 , 7 , 8 on the number

line.

Sol. (i) 6 =

22

2

16

2

16



=

22

2

5

2

7

= 22 )5.2()5.3(

Steps of construction :

(a) Take BC = 2.5 cm.

(b) At C, draw a ray BX making an angle of

90º.

(c) With centre B and radius 3.5 cm draw an

arc which intersects CX at A.

(d) Join AB.

Now, AC = 22 BCAB

= 22 )5.2()5.3(

= 25.625.12 = 6

X

A

CB

3.5

cm

2.5 cm

6 c

m

(ii) 7 =

22

2

17

2

17

=

22

2

6

2

8

= (4)2 – (3)2


(a) Draw BC = 3 cm.

(b) At C, draw a ray CX making an angle of

90º.

(c) With centre B and radius 4 cm, draw an


(d) Join AB.

Now AC = 22 BCAB

= 22 34

= 916 = 7

X

A

CB

4 cm

5 cm

7 c

m

(iii) 8 =

22

2

18

2

18

=

22

2

7

2

9

= 22 )5.3()5.4(


(a) Draw a line segment BC = 3.5 cm.


90º.

X

A

CB

4.5

cm

3.5 cm

8 c

m



(c) With centre B and radius 4.5 cm, draw an


(d) Join AB.

AC = 22 BCAB = 22 )5.3()5.4(

= 5.1225.20 = 8

4. Represent 5.3 , 4.9 and 5.10 on the

real number line.

Sol. (i) 5.3

=

22

2

15.3

2

15.3

=

22

2

5.2

2

5.4

= 22 )25.1()25.2(




90º



(d) Join AB.

AC = 22 BCAB = 22 )25.1()25.2(

= 5625.10625.5 = 5.3

X

A

CB

2.25 cm

1.25 cm

3.5

(ii) 4.9

=

22

2

14.9

2

14.9

=

22

2

4.8

2

4.10

= 22 )2.4()2.5(




90º.

(c) With centre B and radius 5.2 cm draw an


(d) Join AB.

AC = 22 BCAB = 22 )2.4()2.5(

= 64.1704.27 = 40.9 = 4.9

X

A

CB

5.2

cm

4.2 cm

9.4

(iii) 5.10

=

22

2

15.10

2

15.10

=

22

2

5.9

2

5.11

= 22 )75.4()75.5(




90º.


arc intersecting CX at A.



(d) Join AB.

AB = 22 )75.4()75.5(

= 5625.220625.33

= 5000.10 = 5.10

X

A

CB

5.75

cm

4.75 cm10.5

EXERCISE 1.6

1. Visualise 2.665 on the number line, using

successive magnification.

Sol. 2.665

_ It lies between 2 and 3

0 1 2 3 4 5 61

2.660

A A1

2.6 2.7

B B1

2.61

2.61

2.62

2.63

2.64

2.65

2.66

2.67

2.68

2.69

2.67

2.6702.6692.660 2.661 2.662 2.663 2.664 2.665 2.666 2.667 2.668

C C1

2. Visualise the representation of 73.5 on the

number line upto 5 decimal places, that is

upto 5.37777. [NCERT]

Sol. 5.37777

0 1 2 3 4 5 61

5.1

7

65

5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

8

Multiple Choice Questions (MCQs)

Mark the correct alternative in each of the

following:

1. Which one of the following is a correct

statement?

(a) Decimal expansion of a rational number is

terminating

(b) Decimal expansion of a rational number is

non-terminating

(c) Decimal expansion of an irrational number

is terminating

(d) Decimal expansion of an irrational number

is non-terminating and non-repeating

Sol. Decimal expansion of an irrational number

is non-terminating and non-repeating (d)

2. Which one of the following statements is

true?

(a) The sum of two irrational numbers is always

an irrational number

(b) The sum of two irrational numbers is always

a rational number

(c) The sum of two irrational numbers may be

a rational number or an irrational number

(d) The sum of two irrational numbers is always

an integer

Sol. The sum of two irrational numbers may be



a rational number or an irrational number

(c)

3. Which of the following is a correct

statement?

(a) Sum of two irrational numbers is always

irrational

(b) Sum of a rational and irrational number is

always an irrational number

(c) Square of an irrational number is always a

rational number

(d) Sum of two rational numbers can never be

an integer

Sol. Sum of a rational and irrational number is

always an irrational number (b)

4. Which of the following statements is true?

(a) Product of two irrational numbers is always

irrational

(b) Product of a rational and an irrational

number is always irrational

(c) Sum of two irrational numbers can never

be irrational

(d) Sum of an integer and a rational number

can never be an integer

Sol. Product of a rational and an irrational

number is always irrational (b)

5. Which of the following is irrational?

(a)9

4(b)

5

4

(c) 7 (d) 81

Sol. 7 is irrational number

_

9

4 =

3

2 and 81 = 9 (c)


(a) 0.14 (b) 1614.0

(c) 1416.0 (d) 0.1014001400014

Sol. 0.1014001400014..... is irrational as it is

non-terminating nor repeating decimal. (d)

7. Which of the following is rational?

(a) 3 (b)

(c)0

4(d)

4

0

Sol. _ 3 , are irrational and 0

4 is meaningless

4

0 is a rational which is equal to 0. (d)

8. The number 0.318564318564318564..... is:

(a) a natural number

(b) an integer

(c) a rational number

(d) an irrational number

Sol. The number = 0.318564318564318564.....

= 318564.0

_ The decimal is non-terminating and recurring

It is rational number. (c)

9. If n is a natural number, then n is

(a) always a natural number

(b) always a rational number

(c) always an irrational number

(d) sometimes a natural number and sometimes

an irrational number

Sol. If n is a natural number then n may

sometimes a natural number and sometime

an irrational number e.g.

If n = 2 then n = 2 which is are

irrational and if n = 4, then n = 4 = 2

which is a rational number. (d)

10. Which of the following numbers can be

represented as non-terminating, repeating

decimals?

(a)24

39(b)

16

3

(c)11

3(d)

25

137



Sol.11

3 as its denominator has no factor of 2

or 5 or both, so it has non-terminating nor

repeating.

In 24

39 =

8

13,

16

3,

25

137, all have

terminating decimals. (c)

11. Every point on a number line represents

(a) a unique real number

(b) a natural number

(c) a rational number

(d) an irrational number

Sol. Every point on a number line represents a

unique real number. (a)


(a) 0.15 (b) 0.01516

(c) 1516.0 (d) 0.5015001500015..

Sol. As it is non-terminating non-repeating

decimals while others are terminating or

non-terminating repeating decimals. (d)

13. The number 27.1 in the form q

p, where p

and q are integers and q 0, is

(a)9

14(b)

11

14

(c)13

14(d)

15

14

Sol. Let x = 27.1 = 1.272727.....

100x = 127.272727.....

Subtracting, 99x = 126 x = 99

126 =

11

14

(b)

14. The number 3.0 in the form q

p, where p

and q are integers and q 0, is

(a)100

33(b)

10

3

(c)3

1(d)

100

3

Sol. 3.0 = 9

3 =

3

1(c)

15. 23.0 when expressed in the form q

p (p, q

are integers q 0), is

(a)25

8(b)

90

29

(c)99

32(d)

199

32

Sol. 23.0

Let x = 23.0 = 0.3222.....

10x = 3.222.....

100x = 32.2222.....

Subtracting,

90x = 29 x = 90

29

23.0 = 90

29(b)


p(p, q

are integers q 0), is

(a)99

2320(b)

100

2343

(c)999

2343(d)

199

2320

Sol. 43.23

Let x = 43.23 = 23.434343.....

100x = 2343.434343.....

Subtracting,

99x = 2320 x = 99

2320(a)




p(p, q

are integers, q 0), is

(a)1000

1(b)

100

1

(c)1999

1(d)

999

1

Sol. 001.0

Let x = 001.0 = 0.001001001......

1000x = 1.001001001.....

Subtracting,

999x = 1

x = 999

1(d)

18. The value of 23.0 + 22.0 is

(a) 45.0 (b) 43.0

(c) 45.0 (d) 0.45

Sol. 23.0 + 22.0

0.232323..... + 0.222222.....

= 0.454545..... = 45.0 (a)

19. An irrational number between 2 and 2.5 is

(a) 11 (b) 5

(c) 5.22 (d) 5.12

Sol. An irrational number between 2 and 2.5 is

5 as it has approximate value 2.236...

(b)

20. The number of consecutive zeros in 23 × 34

× 54 × 7, is

(a) 3 (b) 2

(c) 4 (d) 5

Sol. In 23 × 34 × 54 × 7, number of consecutive

zero will be 3 as 23 × 54 = 2 × 2 × 2 × 5 ×

5 × 5 × 5 = 5000 (a)

21. The smallest rational number by which 3

1

should be multiplied so that its decimal

expansion terminates after one place of

decimal, is

(a)10

1(b)

10

3

(c) 3 (d) 30

Sol. If 3

1 be multiplied by

10

3, then we get

3

1 ×

10

3 =

10

1 = 0.1 (a)




For any real number a and na positive integer n we define an as

a × a × a × a ..... × a (n times), then

an is called nth power of a.

a is called the base and n is called index, or exponent of nth power of a.

Note = a0 = 1

(i) am × an = am + n (ii) am an = am – n, m > n

(iii) a–m = ma

1(iv) (am)n = amn

(v) (ab)m = am.bm (vi)

m

b

a

= m

m

b

a

(vii) a0 = 1

If a is a positive real number and n is positive integer, then nth root of a is denoted by na

1

and

is written as n a

Note : n

m

a = n ma

EXERCISE 2.1

1. Simplify the following:

(i) 3 (a4b3)10 × 5(a2b2)3 (ii) (2x–2y3)3 (iii) 4

57

108

)106()104(

(iv) 22

32

10

)5(4

ba

abab (v)

n

ba

yx

32

22

(vi) 42

693 )(

n

n

a

a

Sol. (i) 3 (a4b3)10 × 5(a2b2)3

= 3a4 × 10 b3 × 10 × 5a2 × 3 × b2 × 3

nmnm

mnnm

aaa

aa )(

= 3a40b30 × 5a6b6

= 3 × 5 × a40 + 6 × b30 + 6

= 15a46b36

(ii) (2x–2y3)3 = 23 × x–2 × 3 × y3 × 3

nmnm

mnnm

aaa

aa )(

EXPONENTS OF REAL NUMBERS2



= 8 × x–6 × y9 = 8x–6y9

(iii) 4

57

108

)106()104(

= 8

64 107 – 5 – 4

10a

aaa nmnm

= 3 10–2 = 210

3 =

1010

3

= 100

3

(iv) 22

32

10

)5(4

ba

abab =

10

)5(4 × a1 + 1 – 2 b2 + 3 – 2

10a

aaa nmnm

= –2 × a0b3 = –2 × 1 × b3

= –2b3

(v)

n

ba

yx

32

22

= nn

nn

ba

yx32

22 = nn

nn

ba

yx32

22 ·{_ (am)n = amn}

(vi) 42

693 )(

n

n

a

a = 42

6)93(

n

n

a

a

mnnm

mnnm

aa

aaa

)(

= 42

5418

n

n

a

a = a18n – 54 – 2n + 4 = a16n – 50

2. If a = 3 and b = –2, find the values of:

(i) aa + bb (ii) ab + ba (iii) (a + b)ab

Sol. a = 3 and b = –2

(i) aa + bb = a3 + b–2

= (3)3 + (–2)–2

= 27 + 2)2(

1

= 27 + )2(2

1

= 27 + 4

1 =

4

1108 =

4

109

(ii) ab + ba = (3)–2 + (–2)3

= 23

1 + (–2) × (–2) × (–2)

= 33

1

+ (–8) =

9

1 – 8 =

9

721 =

9

71

(iii) (a + b)ab = [3 + (–2)]3 × (–2)

= (3 – 2)–6 = (1)–6 = 6)1(

1 =

1

1 = 1



3. Prove that:

(i)

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x

= 1

(ii)

c

b

a

x

x

×

a

c

b

x

x

×

b

a

c

x

x

= 1

(iii)

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x

= 1

Sol. (i)

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x

= 1

LHS =

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x

= 22

)( bababax × 22

)( cbcbcbx × 22

)( acacacx

= )()(22

bababax × )()(22

cbcbcbx × )()(22

acacacx

= 33

bax · 33

cbx · 33

acx

= 333333

accbbax = x0 = 1 = RHS

(ii)

c

b

a

x

x

×

a

c

b

x

x

×

b

a

c

x

x

= 1

LHS =

c

b

a

x

x

×

a

c

b

x

x

×

b

a

c

x

x

= (xa – b)c × (xb – c)a × (xc – a)b

= x(a – b)c × x(b – c)a × x(c – a)b

= xac – bc × xab – ac × xbc – ab

= xac – bc + ab – ac + bc – ab = x0 = 1 = RHS

(iii)

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x

= 1

LHS =

22 baba

b

a

x

x

×

22 cbcb

c

b

x

x

×

22 acac

a

c

x

x



= 22

)( bababax × 22

)( cbcbcbx × 22

)( acacacx

= )()(22

bababax × )()(22

cbcbcbx × )()(22

acacacx

= 33

bax · 33

cbx · 33

acx

= 333333

accbbax = )(2333

cbax

4. Prove that:

(i) bax 1

1 + abx 1

1 = 1

(ii) acab xx 1

1 + bcba xx 1

1 + cacb xx 1

1 = 1

Sol. (i) bax 1

1 + abx 1

1 = 1

LHS = bax 1

1 + abx 1

1

= babb xx

1 + abaa xx

1{Put 1 = xb – b and 1 = xa – a}

= )(

1abb xxx +

)(

1baa xxx

= ba

b

xx

x

+ ba

a

xx

x

= )( ba

ab

xx

xx

=

)( ba

ba

xx

xx

= 1 = RHS

(ii) acab xx 1

1 + bcba xx 1

1 + cacb xx 1

1 = 1

LHS = acab xx 1

1 + bcba xx 1

1 + cacb xx 1

1

= acabaa xxx

1 + bcbabb xxx

1 + cacbcc xxx

1

{Put 1 = xa – a, 1 = xb – b and 1 = xc – c}

= )(

1cbaa xxxx +

)(

1cabb xxxx +

)(

1abcc xxxx



= cba

a

xxx

x

+ cba

b

xxx

x

+ cba

c

xxx

x

= cba

cba

xxx

xxx

= 1 = RHS

5. Prove that:

(i) 111111

accbba

cba = abc (ii) (a–1 + b–1)–1 =

ba

ab

Sol. (i) 111111

accbba

cba = abc

LHS = 111111

accbba

cba

m

m

aa

1

=

cabcab

cba

111

=

abc

bac

cba

= )(

)(

cba

abccba

= abc = RHS

(ii) (a–1 + b–1)–1 = ba

ab

LHS = (a–1 + b–1)–1

=

111

ba =

1

ab

ab

m

m

aa

1

= ba

ab

= RHS

6. If abc = 1, show that 11

1 ba

+ 11

1 cb

+ 11

1 ac

= 1

Sol. abc = 1

11

1 ba

+ 11

1 cb

+ 11

1 ac

= 1

LHS = 11

1 ba

+ 11

1 cb

+ 11

1 ac

= b

a1

1

1

+ c

b1

1

1

+ a

c1

1

1



=

b

abb 1

1

+ abb 1

1 +

aab

111

1

abc

abc

abc

1and

1

1

= 1 abb

b +

abb 1

1 +

bab

ab

1

= abb

abb

1

1 = 1

7. Simplify the following:

(i) 11

1

93

93

nn

nn(ii) 132

21

)25(55

525255

nn

nn

(iii) nn

nn

5259

5652

13

(iv) nn

nn

)8(7)2(10

)2(16)8(613

231

Sol. (i) 11

1

93

93

nn

nn = 121

12

)3(3

)3(3

nn

nn

= 221

22

33

33

nn

nn

= 32n + 2 + n – n + 1 – 2n + 2

= 33n – 3n + 2 + 3 = 35 = 3 × 3 × 3 × 3 × 3 = 243

(ii) 132

21

)25(55

525255

nn

nn

= 1232

2212

)5(55

55)5(5

nn

nn

= 2232

2222

55555

55555

nn

nn

= )555(5

)555(5232

222

n

n

= 24

23

55

55

=

25625

25125

= 600

100 =

6

1

(iii) nn

nn

5259

5652

13

= )29(5

)565(52

13

n

n

= 49

30125

= 5

95 = 19

(iv) nn

nn

)8(7)2(10

)2(16)8(613

231

= nn

nn

)2(7)2(10

)2(16)2(6313

2313

= nn

nn

313

2333

27210

21626

= )7210(2

)21626(213

233

n

n



= 720

4

11648

=

13

448 =

13

52 = 4

8. Solve the following equations for x:

(i) 72x + 3 = 1 (ii) 2x + 1 = 4x – 3

(iii) 25x + 3 = 8x + 3 (iv) 42x = 32

1

(v) 4x – 1 × (0.5)3 – 2x =

x

8

1

(vi) 23x – 7 = 256

Sol. (i) 72x + 3 = 1 = 70

(_ a0 = 1)

Comparing, we get

2x + 3 = 0 2x = –3

x = 2

3

(ii) 2x + 1 = 4x – 3

2x + 1 = (22)x – 3

2x + 1 = 22x – 6

Comparing, we get

x + 1 = 2x – 6

1 + 6 = 2x – x x = 7

x = 7

(iii) 25x + 3 = 8x + 3

25x + 3 = (23)x + 3

25x + 3 = 23x + 9

Comparing, we get

5x + 3 = 3x + 9

5x – 3x = 9 – 3

3x = 6 x = 3

6 = 2

x = 2

(iv) 42x = 32

1 (22)2x = 52

1

24x = 2–5

Comparing, we get

4x = –5 x = 4

5

x = 4

5

(v) 4x – 1 × (0.5)3 – 2x =

x

8

1

(22)x – 1 ×

x23

2

1

=

x

32

1

22x – 2 × 2–3 + 2x = 2–3x

22x – 2 – 3 + 2x = 2–3x

24x – 5 = 2–3x

Comparing, we get

4x – 5 = –3x 4x + 3x = 5

7x = 5 x = 7

5

x = 7

5

(vi) 23x – 7 = 256

23x – 7 = 28

Comparing, we get

3x – 7 = 8

3x = 8 + 7 = 15

x = 3

15 = 5

x = 5

9. Solve the following equations for x:

(i) 22x – 2x + 3 + 24 = 0 (ii) 32x + 4 +

1 = 2 · 3x + 2

Sol. (i) 22x – 2x + 3 + 24 = 0

(2x)2 – 23 × (2)x + 24 = 0

(2x)2 – 8(2)x + 16 = 0

(2x)2 – 2 × 2x × 4 + (4)2 = 0

(2x – 4)2 = 0 2x – 4 = 0

2x = 4 = 22

Comparing, we get

x = 2

(ii) 32x + 4 + 1 = 2 · 3x + 2

32x × 34 + 1 – 2 · 3x × 32 = 0

81 · 32x – 2 × 9(3x) + 1 = 0



81(3x)2 – 18(3x) + 1 = 0

[9(3x)]2 – 2 × 9 × 3x + (1)2 = 0

[9(3x) – 1]2 = 0

9(3)x – 1 = 0 9(3x) = 1

3x = 9

1 = 23

1 = 3–2

Comparing, we get

x = –2

10. If 49392 = a4b2c3, find the values of a, b

and c, where a, b and c are different positive

primes.

Sol. 49392 = 2a × 3b × 7c

24 × 32 × 73 = a4b2c3

2 493922 246962 123482 61743 30873 10297 3437 497 7

1

Comparing, we get

a = 4, b = 2, c = 7

11. If 1176 = 2a × 3b × 7c, find a, b and c.

Sol. 1176 = 2a × 3b × 7c

23 × 31 × 72 = 2a × 3b × 7c

2 11762 5882 2943 1477 497 7

1

Comparing, we get

a = 3, b = 1 = c = 2

12. Given 4725 = 3a 5b 7c, find:

(i) the integral values of a, b and c

(ii) the value of 2–a 3b 7c

Sol. 4725 = 3a · 5b · 7c

(i) 33 · 52 · 71 = 3a · 5b · 7c

Comparing, we get

a = 3, b = 2, c = 1

(ii) 2–a 3b 7c = 2–3 · 32 · 71

= 32

1 × 32 × 71

= 222

733

= 8

63

13. If a = xyp – 1, b = xyq – 1 and c = xyr – 1, prove

that aq – r br – p cp – q = 1.

Sol. a = xyp – 1, b = xyq – 1 and c = xyr – 1

aq – r br – p cp – q = 1

LHS = aq – r br – p cp – q

= (xyp – 1)q – r · (xyq – 1)r – p · (xyr – 1)p – q

= xq – r · y(p – 1) (q – r) · xr – p · y(q – 1) (r – p) · xp –

q · y(r – 1) (p – q)

= xq – r + r – p + p – q · ypq – pr – q – pr · yqr – pq – r + p ·

yrp – rq – p + q

= x0 · ypq – pr – q + r + qr – pq – r + p + rp – rq – p + q

= x0 · y0 = 1 × 1 = 1 = RHS

nmnm

mnnm

aaa

aa

a

)(

10

EXERCISE 2.2

1. Assuming that x, y, z are positive real

numbers, simplify each of the following:

(i)5

3

x (ii) 23 yx

(iii) (x–2/3y–1/2)2

(iv) 3/2)( x 4y 2/1xy

(v) 5 10510243 zyx

(vi)

4/5

10

4

y

x(vii)

5

3

2

2

7

6



Sol. (i) 5

3

x =

5

2

3

x = 5

2

3

x

= 2

15

x =

2

15

1

x

m

m

mnnm

xx

xx

1

)(

(ii) 23 yx = 2

123 yx

= 2

3

x. 2

2

y = 2

3

x.y–1

m

m

mnnm

xx

xx

1and

)(

= y

x 2

3

(iii)

2

2

1

3

2

yx = 23

2

x.

22

1

y

= 3

4

x.y–1 =

yx 3

4

1

m

m

mnnm

xx

xx

1

)(

(iv) 3/2)( x 4y 2/1xy

=

3

2

2

1

x .

4

2

1

y 2

1

x

2

1

2

1

y

=

3

2

2

1

x.

42

1

y 2

1

x. 2

1

2

1

y

= 3

1

x.y2 2

1

x. 4

1

y

=

2

1

3

1

x.

4

12

y

= 6

32

x. 4

12

y = 6

5

x. 4

9

y

=

6

5

4

9

x

y

(v) 5 10510243 zyx = 5

110510 ..243 zyx

= 5

1105105 ...3 zyx

= 5

15

3 . 5

110

x. 5

15

y . 5

110

z = 3.x2yz2

(vi)

4/5

10

4

y

x =

4

5

4

10

x

y

m

m

xx

1

=

4

54

4

510

x

y = 5

2

25

x

y

(vii)

5

3

2

2

7

6

=

5

3

2

×

2

7

6

=

5

3

2

× 2

2

)7(

)6(

= 2

5

3

×

49

36

=

2

3

2

×

3

2 ×

49

36

= 9

11 ×

3

2 ×

49

36

= 349

216 =

49493

16162

= 7203

512 =

3

1

7203

512



2. Simplify:

(i) (16–1/5)5/2 (ii) 5 3)32(

(iii) 3 2)343( (iv) (0.001)1/3

(v) 3/44/5

5/32/3

)8()16(

)243()25(

(vi)

8

5

2

13

5

2

(vii)2

7

42

21

75

75

× 2

5

53

32

75

75

Sol. (i) (16–1/5)5/2 = 2

5

5

142

= 2

5

5

14

2

m

m

mnnm

aa

aa

1

)(

= 2–2 = 22

1

= 22

1

=

4

1

(ii) 5 3)32( = 5332

1 =

5

3

32

1 =

5

3

5 )2(

1

=

5

35

2

1

= 32

1 =

222

1

=

8

1

(iii) 3 2)343( = 32343

1 =

3

2

343

1

=

3

1

3 )7(

1 =

3

23

7

1

= 27

1 =

77

1

=

49

1

(iv) (0.001)1/3 = 3

13)1.0( = 3

13

)1.0(

= (0.1)1 = 0.1

(v) 3/44/5

5/32/3

)8()16(

)243()25(

=

3

4

34

5

4

5

3

52

3

2

)2()2(

)3()5(

=

3

43

4

54

5

35

2

32

22

35

= 45

33

22

35

= 222222222

333555

= 512

3375

(vi)

8

5

2

13

5

2

=

138

5

2

=

5

5

2

= 5

5

5

)2(

= 5

5

)2(

5

= 2

5

5

2

5

= 22222

55555

= 222

3125

=

24

3125

(vi)2

7

42

21

75

75

× 2

5

53

32

75

75

=

2

74

2

72

2

72

2

71

75

75

×

2

55

2

53

2

53

2

52

75

75

= 147

72

7

75

75

=

2

25

2

15

2

15

5

75

75



= 2

1575

2

7

5

× 2

2514

2

157

7

= 2

21

2

25

5 . 2

4021

7 = 2

4

5. 2

1

7

= 52 × 71 = 25 × 7 = 175

3. Prove that:

(i) 353 3 13 5 × 6 653 = 5

3

(ii) 2

3

9 – 3 × 50 – 2

1

81

1

= 15

(iii)

2

4

1

– 3 × 3

2

8 × 40 + 2

1

16

9

= 3

16

(iv)

5

3

5

1

4

1

3

1

2

1

510

432

64

53

5

3

5

7

3

4

= 10

(v)4

1 + (0.01)–1/2 – (27)2/3 =

2

3

(vi) nn

nn

22

221

1

= 2

3

(vii)3

2

125

64

+ 4

1

625

256

1

+

0

3 64

25

=

16

61

(viii) 3/13/432

23

3)15(25/15

9863

= 28 2

(ix)131

10

3

1

2

3

8

3

)1.0()6.0(

= –

2

3

Sol. (i) 353 3 13 5 × 6 653 = 5

3

L.H.S. = 353 3 13 5 × 6 653

= 2

1353 3

113 2

1

5 × 6

1653

=

2

3

2

1

53

2

1

3

1

53 ×

6

16

6

1

53

=

2

3

2

1

53

2

1

3

1

53 × 16

1

53

= 6

1

3

1

2

1

3

. 12

1

2

3

5

= 6

1

3

1

2

1

3 . 2

213

5

= 6

123

3

. 2

24

5

= 6

6

3 × 2

2

5

= 31.5–1

= 5

3 = R.H.S.

(ii) 2

3

9 – 3 × 50 – 2

1

81

1

= 15

L.H.S. = 2

3

9 – 3 × 50 – 2

1

81

1

= 2

323 – 3 × 1 –

2

1

29

1

{_ a0 = 1}

= 2

32

3

– 3 –

2

12

9

1

= 33 – 3 –

1

9

1

= 27 – 3 –

1

9

1

= 27 – 3 – 9 = 15 = R.H.S.



(iii)

2

4

1

– 3 × 3

2

8 × 40 + 2

1

16

9

= 3

16

L.H.S. =

2

4

1

– 3 × 3

2

8 × 40 + 2

1

16

9

=

22

2

1

– 3 × 3

232 × 1 +

2

12

4

3

=

)2(2

2

1

– 3 × 3

23

2

× 1 +

2

12

4

3

=

4

2

1

– 3 × 22 × 1 +

1

4

3

=

4

1

2

– 3

× 4 + 3

4

m

m

aa

1

=

4

1

2

– 12 + 3

4 = 16 – 12 +

3

4

= 4 + 3

4 =

3

16 = R.H.S.

(iv)

5

3

5

1

4

1

3

1

2

1

510

432

64

53

5

3

5

7

3

4

= 10

L.H.S. =

5

3

5

1

4

1

3

1

2

1

510

432

64

53

5

3

5

7

3

4

=

5

3

5

1

4

1

23

1

2

1

5)52(

)2(32

115

3

2

5

7

3

4

32)2(

53

=

5

3

5

1

5

1

4

12

3

1

2

1

552

232

115

6

5

7

3

4

322

53

=

5

3

5

1

5

1

2

1

3

1

2

1

552

232

×

5

7

3

4

115

6

53

322

= 5

11

5

6

2

1

2

1

2 . 3

41

3

1

3 . 5

7

5

3

5

1

5

= 10

2101255

2

. 3

431

3

. 5

731

5

= 10

1222

2

.34 – 4. 5

38

5

= 10

10

2.30. 5

5

5

= 21.30.51 = 2 × 1 × 5 = 10

= R.H.S.

(v)4

1 + 2

1

)01.0(

– 3

2

)27( = 2

3

L.H.S. = 4

1 + 2

1

)01.0(

– 3

2

)27(

= 2

1

22

1

+

2

12

)1.0( – 3

2

3 )3(

= 2

1

2 )2(

1

+

2

12

)1.0( – 3

23

3

= 12

1 + (0.1)–1 – 32

= 2

1 +

1

1

10

– 32

= 2

1 + 10 – 9 = 1

2

1 =

2

3 = R.H.S.

(vi) nn

nn

22

221

1

= 2

3

L.H.S. = nn

nn

22

221

1

= nn

nn

22.2

2221

1

= )12(2

)21(21

1

n

n

= 12

2

11

= 1

2

3



= 2

3 = R.H.S.

(vii)3

2

125

64

+ 4

1

625

256

1

+

0

3 64

25

=

16

61

L.H.S. = 3

2

125

64

+ 4

1

625

256

1

+

3 64

25

= 3

2

3

3

5

4

+

4

1

4

4

5

4

1

+

3 3

2

4

5

=

3

25

4

3

3

23

+

4

14

4

14

5

4

1

+

3

1

3

2

1

2

)4(

)5(

= 2

2

5

4

+

5

4

1 +

4

5

= 2

2

4

5 +

4

5 +

4

5

= 16

25 +

4

10 =

16

25 +

16

40

= 16

65 = R.H.S.

(viii)

3

1

3

4

32

23

3)15(25

15

9863

= 28 2

L.H.S. =

3

1

3

4

32

2

1

23

3)15(25

15

)98(63

=

3

1

3

4

3

4

3

1

2

2

1

223

353)25(5

)492()32(3

=

3

1

3

4

3

4

3

1

22

2

1

22

1

223

353)5(5

)7(2323

= 22. 2

1

2. 3

1

3

423

3 . 3

42

3

2

5 .71

= 4 2 × 30.50 × 71

= 4 2 × 1 × 1 × 7

= 28 2 = RHS

(ix)131

10

3

1

2

3

8

3

)1.0()6.0(

= –

2

3

L.H.S. = 131

10

3

1

2

3

8

3

)1.0()6.0(

= 131

1

1

3

2

3

3

8

10

11

=

1

3

8

27

3

8

101 1

=

1

39

9

= 6

9 =

2

3 = RHS

4. Show that:

(i) bax 1

1 + abx 1

1 = 1



(ii)

ba

abb

abb

baa

baa

x

x

x

x

)(

)(

)(

)(

= 1

(iii)

ca

bax

1

1 ab

cbx

1

1 bc

acx

1

1

= 1

(iv)

ba

ab

ba

x

x

22 cb

bc

cb

x

x

22 ca

ac

ac

x

x

22

=

)(2333

cbax

(v) (xa – b)a + b (xb – c)b + c (xc – a)c + a = 1

(vi) 1

1

11

a

a

aaax = x

(vii)

yx

y

x

a

a

1

1zy

z

y

a

a

2

2xz

x

z

a

a

3

3

= 1

(viii)

ba

b

a

3

3cb

c

b

3

3ac

a

c

3

3 = 1

Sol. (i) bax 1

1 + abx 1

1 = 1

LHS = bax 1

1 + abx 1

1

= babb xx

1 + abaa xx

1

= )(

1abb xxx +

)(

1baa xxx

= ba

b

xx

x

+ ba

a

xx

x

= ba

ab

xx

xx

= ba

ba

xx

xx

= 1 = RHS.

(ii)

ba

abb

abb

baa

baa

x

x

x

x

)(

)(

)(

)(

= 1

LHS =

ba

abb

abb

baa

baa

x

x

x

x

)(

)(

)(

)(

=

ba

abb

abb

aba

aba

x

x

x

x

2

2

2

2

= baabbabbabaaba

xx

2222

= [x–2ab ÷ x–2ab]a + b

= [x–2ab – (–2a)]a + b

= (x–2ab + 2ab)a + b = (x0)a + b = (x)0 = 1 = RHS

(iii)

ca

bax

1

1 ab

cbx

1

1 bc

acx

1

1

= 1

LHS =

ca

bax

1

1 ab

cbx

1

1 bc

acx

1

1

= )(

1

)(

1

cabax

× abcbx

11

× bcacx

11

= ))((

1

)()(

1

)()(

1

bcacabcbcabax

= ))((

1

)()(

1

)()(

1

cbacbacbacbax

= )()()( accbba

baaccb

x

= )()()(

0

accbbax = x0 =

1 = RHS

(iv)

ba

ab

ba

x

x

22 cb

bc

cb

x

x

22 ca

ac

ac

x

x

22

= )(2333

cbax

LHS =

ba

ab

ba

x

x

22 cb

bc

cb

x

x

22 ca

ac

ac

x

x

22



= baabbax

22 cb

bccbx

22 cacaacx

22

= )()( 22 bababax ·

)()( 22 cbcbcbx

·)()(

22cacaacx

= 33

bax ·

33cbx

·33

acx

= 333333

accbbax =

)(2333

cbax = RHS

(v) (xa – b)a + b (xb – c)b + c (xc – a)c + a = 1

LHS = (xa – b)a + b (xb – c)b + c (xc – a)c + a

= x(a – b) (a + b) · x(b – c) (b + c) · x(c – a) (c + a)

= 22

bax ·22

cbx ·22

acx

= 222222

accbbax = x0 = 1 = RHS

(vi) 1

1

11

a

a

aaax = x

LHS = 1

1

11

a

a

aaax

=

1

1

1

1

a

a

aa

a

x = 11

112

a

a

aa

a

x

= 11

1)1()1(

a

a

aa

aa

x = x1 = x = RHS

(vii)

yx

y

x

a

a

1

1zy

z

y

a

a

2

2xz

x

z

a

a

3

3

= 1

LHS =

yx

y

x

a

a

1

1zy

z

y

a

a

2

2xz

x

z

a

a

3

3

= (ax + 1 – y – 1)x + y × (ay + 2 – z – 2)y + z

.(az + 3 – x – 3)z + x

= (ax – y)x + y · (ay – z)y + z · (az – x)z + x

= a(x – y) (x + y) · a(y – z) (y + z) · a(z – x) (z + x)

= 22

yxa ·22

zya ·22

xza

= 222222

xzzyyxa = a0 = 1 = RHS

(viii)

ba

b

a

3

3cb

c

b

3

3ac

a

c

3

3 = 1

LHS =

ba

b

a

3

3cb

c

b

3

3ac

a

c

3

3

= (3a – b)a + b · (3b – c)b + c · (3c – a)c + a

= 22

3 ba ·22

3 cb ·22

3 ac

= 222222

3 accbba = 30 = 1 = RHS

5. If 2x = 3y = 12z, show that z

1 =

y

1 +

x

2.

Sol. 2x = 3y = 12z, z

1 =

y

1 +

x

2

Let 2x = 3y = 12z = k

Then, 2 = xk

1, 3 = yk

1

, 12 = zk

1

22 × 3 = 3

1

k

2

1

)( xk × )(

1

yk = zk

1

xk

2

· yk

1

= 3

1

k

yxk

12

= zk

1

Comparing both sides,

z

1 =

y

1 +

x

2Hence Proved.

6. If 2x = 3y = 6–z, show that x

1 +

y

1 +

z

1 =

0.

Sol. Let 2x = 3y = 6–z = k, then



2 = xk

1

, 3 = yk

1

and 6 = zk

1

2 × 3 = zk

1

xk

1

× yk

1

= zk

1

yxk

11

= zk

1

x

1 +

y

1 =

z

1

x

1 +

y

1 +

z

1 = 0 Hence Proved.

7. If ax = by = cz and b2 = ac, then show that

y = xz

zx

2

.

Sol. ax = by = cz = k, then

a = xk

1

, b = yk

1

and c = zk

1

Now, b2 = ac

2

1

)( yk = xk

1

· zk

1

yk

2

= zxk

11

y

2 =

x

1 +

z

1

y

2 =

xy

xz

y = xz

xy

2

8. If 3x = 5y = (75)z, show that z = yx

xy

2.

Sol. Let 3x = 5y = 75z = k

3 = xk

1

, 5 = yk

1

and 75 = zk

1

75 = zk

1

3 × 52 = zk

1

xk

1

× yk

2

= zk

1

yxk

21

= zk

1

x

1 +

y

2 =

z

1

xy

xy 2 =

z

1

z = yx

xy

2

9. If 27x = x3

9, find x.

Sol. 27x = x3

9 (33)x = x

3

32

33x × 3x = 32 33x + x = 32

34x = 32

Comparing, we get

4x = 2 x = 4

2 =

2

1

x = 2

1

10. Find the values of x in each of the following:

(i) 25x 2x = 5 202 (ii) (23)4 = (22)x

(iii)27

125

3

5

5

32

xx

(iv) 5x – 2 × 32x – 3 = 135

(v) 2x – 7 × 5x – 4 = 1250 (vi) 2

12

3 )4(x

= 32

1

(vii) 52x + 3 = 1 (viii)x)13( = 44 – 34 – 6

(ix)

1

5

3

x

= 27

125



Sol. (i) 25x 2x = 5 202

x

x

2

25

= 5

1

20 )2( 25x – x = 5

20

2

24x = 5

20

2

Comparing, we get

4x = 5

20 x =

45

20

= 1

x = 1

(ii) (23)4 = (22)x

23 × 4 = 22 × x 212 = 22x

Comparing, we get

2x = 12 x = 2

12 = 6

x = 6

(iii)27

125

3

5.

5

32

xx

x

3

5.

x2

3

5

= 3

3

3

5

xx 2

3

5

=

3

3

5

Comparing, we get

–x + 2x = 3 x = 3

x = 3

(iv) 5x – 2 × 32x – 3 = 135

5x.5–2.32x.3–3 = 135

32

2

35

3.5

xx

= 135

5x.32x = 135 × 52 × 33

5x × 3x × 3x = 135 × 25 × 27

(5 × 3 × 3)x = 3 × 3 × 3 × 5 × 5 × 5 × 3 ×

3 × 3

(45)x = (3 × 5 × 3)3 = (45)3

Comparing, we get

x = 3

(v) 2x – 7 × 5x – 4 = 1250

2x.2–7.5x.5–4 = 2 × 5 × 5 × 5 × 5

4752

52

xx

= 2 × 5 × 5 × 5 × 5

(10)x = 2 × 5 × 5 × 5 × 5 × 2 × 2 × 2 × 2

× 2 × 2 × 2 × 5 × 5 × 5 × 5

(10)x = 28 × 58 = (10)8

Comparing, we get

x = 8

(vi) 2

12

3 )4(x

= 32

1

2

12

3

1

2 )2(

x

= 52

1

2

12

3

2

2x

= 2–5

3

2

2

12x = –5

3

4x +

6

2 = –5 8x + 2 = –5 × 6

8x = –30 – 2 x = 8

32 = –4

(vii) 52x + 3 = 1 = 50

(_ a0 = 1)

2x + 3 = 0 2x = –3

x = 2

3

(viii)x)13( = 44 – 34 – 6

= 256 – 81 – 6 = 169 = (13)2

x = 2

Squaring, x = 4

(ix)

1

5

3

x

= 27

125

1

2

1

5

3

x

=

3

3

5



2

1

5

3

x

=

3

5

3

m

m

aa

1

2

1x = –3

x + 1 = –6 x = –6 – 1 = –7

x = –7

11. If x = 21/3 + 22/3, show that x3 – 6x = 6.

Sol. x = 3

1

2 + 3

2

2 , x3 – 6x = 6

Cubing both sides,

x3 =

3

3

2

3

1

22

= 2 + 22 + 3 × 3

1

2 ×

3

2

2

3

2

3

1

22

x3 = 2 + 4 + 3 × 3

2

3

1

2

× x

x3 = 6 + 3 × 2 × x

x3 = 6x + 6

x3 – 6x = 6 Hence Proved.

12. Determine (8x)x, if 9x + 2 = 240 + 9x.

Sol. (8x)x, if 9x + 2 = 240 + 9x

9x + 2 = 240 + 9x

9x × 92 = 240 + 9x 9x × 81 = 9x + 240

81 × 9x – 9x = 240 80 × 9x = 240

9x = 80

240 = 3 32x = 31

2x = 1 x = 2

1

Now, (8x)x = 2

1

2

18

= 2

1

4 = 2

1

2 )2( =

21 = 2

13. If 3x + 1 = 9x – 2, find the value of 21 + x.

Sol. 3x + 1 = 9x – 2

3x + 1 = (32)x – 2 3x + 1 = 32x – 4

x + 1 = 2x – 4 1 + 4 = 2x – x

x = 5

Now, 21 + x = 21 + 5 = 26 = 64

14. If 34x = (81)–1 and 101/y = 0.0001, find the

value of 2–x + 4y.

Sol. 34x = (81)–1, y

1

10 = 0.0001

34x = (34)–1 34x = 3–4

4x = –4 x = 4

4 = –1, and

y

1

10 = 0.0001 = 10000

1 = 4)10(

1 = (10)–4

y

1 = –4 y =

4

1

Now, 2–x + 4y =

4

14)1(

2

= 21 – 1 = 20 = 1

15. If 53x = 125 and 10y = 0.001 find x and y.

Sol. 53x = 125 and 10y = 0.001

53x = 125 = (5)3

3x = 3 x = 3

3 = 1

and 10y = 0.001 = 1000

1 = 310

1 = 10–3

y = –3

and x = 1, y = –3

16. Solve the following equations:

(i) 3x + 1 = 27 × 34

(ii) 42x = y/63 )16( = 2)8(

(iii) 3x – 1 × 52y – 3 = 225

(iv) 8x + 1 = 16y + 2 and,

x

3

2

1 =

y3

4

1

(v) 4x – 1 × (0.5)3 – 2x =

x

8

1



(vi)b

a =

x

a

b21

, where a, b are distinct

positive primes.

Sol. (i) 3x + 1 = 27 × 34

3x + 1 = 33 + 4 = 37

x + 1 = 7 x = 7 – 1 = 6

x = 6

(ii) 42x = y/63 )16( = 2)8(

42x = 2)8( = 22

1

8 = 8 = (2)3

(22)2x = 23 22 × 2x = 23

24x = 23

Comparing, we get

4x = 3 x = 4

3

and y

6

3 )16(

= 2)8( = 23

y

6

3 4 )2(

= 23

y

6

3

4

2

= 23

y

6

3

4

2 = 23

y

8

2

= 23

Comparing, we get

y

8 = 3 y =

3

8

(iii) 3x – 1 × 52y – 3 = 225

3x – 1 × 52y – 3 = (15)2 = (3 × 5)2

3x – 1 × 52y – 3 = 32 × 52

Comparing,

3x – 1 = 32 x – 1 = 2 x = 2 + 1 = 3

and,

52y – 3 = 52 2y – 3 = 2 2y = 2 + 3 = 5

y = 2

5

x = 3, y = 2

5

(iv) 8x + 1 = 16y + 2 and,

x

3

2

1 =

y3

4

1

(23)x + 1 = (24)y + 2

23x + 3 = 24y + 8

3x + 3 = 4y + 8

3x – 4y = 8 – 3 = 5 ...(i)

and

x

3

2

1 =

y3

4

1

x

3

2

1 =

y32

2

1

=

y6

2

1

3 + x = 6y x = 6y – 3 ...(ii)

From (i),

3(6y – 3) – 4y = 5 18y – 9 – 4y = 5

14y = 5 + 9 = 14 y = 14

14 = 1

and x = 6 × 1 – 3 = 6 – 3 = 3

x = 3, y = 1

(v) 4x – 1 × (0.5)3 – 2x =

x

8

1

(22)x – 1 ×

x23

2

1

=

x

32

1

22x – 2 × 2–3 + 2x = 2–3x

22x – 2 – 3 + 2x = 2–3x 24x – 5 = 2–3x

4x – 5 = –3x 4x + 3x = 5

7x = 5 x = 7

5

(vi)b

a =

x

a

b21

2

1

b

a =

x

b

a21

2

1 = –1 + 2x 2x = 1 +

2

1 =

2

3



x = 22

3

=

4

3

x = 4

3

17. If a and b are distinct positive primes such

that 3 46 ba = axb2y, find x and y.

Sol. 3 46 ba = axb2y

3

1

46 )( ba = ax·b2y

3

6

a · 3

4

b = ax·b2y a2· 3

4

b = ax · b2y

Comparing, we get

ax = az x = 2

and 3

4

b = b2y 2y = 3

4

y = 23

4

= 3

2

x = 2, y = 3

2

18. If a and b are different positive primes

such that

(i)

7

42

21

ba

ba ÷

32

53

ba

ba = axby, find x and y.

(ii) (a + b)–1 (a–1 + b–1) = axby, find x + y + 2.

Sol. (i)

7

42

21

ba

ba ÷

32

53

ba

ba = axby

2814

147

·

ba

ba ÷ 32

53

ba

ba

= axby

2814

147

·

ba

ba × 53

32

ba

ba = axby

a–7 – 14 – 2 – 3 · b14 + 28 + 3 + 5 = axby

a–26 × b50 = axby

Comparing, we get

x = –26, y = 50

(ii) (a + b)–1 (a–1 + b–1) = axby, find x + y + 2

(a + b)–1

ba

11 = axby

ba 1

× ab

ab = axby

ab

1 = axby a–1b–1 = ax·by

Comparing,

x = –1, y = –1

Now put the value of x and y in x + y + 2,

x + y + 2 = –1 – 1 + 2 = 0

19. If 2x × 3y × 5z = 2160, find x, y and z.

Hence, compute the value of 3x × 2–y × 5–z.

Sol. 2x × 3y × 5z = 2160

2 21602 10802 5402 2703 1353 453 155 5

1

2x × 3y × 5z = 24 × 33 × 51

Comparing, we get

x = 4, y = 3, z = 1

Now, 3x × 2–y × 5–z

= 34 × 2–3 × 5–1 = 58

81

=

40

81

20. If 1176 = 2a × 3b × 7c, find the values of a,

b and c. Hence, compute the value of 2a ×

3b × 7–c as a fraction.

Sol. 1176 = 2a × 3b × 7c

2 11762 5882 2943 1477 497 7

1



2a × 3b × 7c = 23 × 31 × 72

Comparing, we get

a = 3, b = 1, c = 2

2a × 3b × 7–c

= 23 × 31 × 7–2 = 8 × 3 × 49

1 =

49

24

21. Simplify:

(i)

ba

c

ba

x

x

cb

a

cb

x

x

ac

b

ac

x

x

(ii)lm

m

l

x

x ×

mn

n

m

x

x ×

nl

l

n

x

x

Sol. (i)

ba

c

ba

x

x

cb

a

cb

x

x

ac

b

ac

x

x

= (xa + b – c)a – b·(xb + c – a)b – c·(xc + a – b)c – a

= x(a – b) (a + b – c)·x(b – c) (b + c – a)·x(c – a) (c + a – b)

= bcacbax 22

·acabcbx 22

·abbcacx 22

= abbcacacabcbbcacbax 222222

= x0 = 1

(ii)lm

m

l

x

x ×

mn

n

m

x

x ×

nl

l

n

x

x

= lmmlx

1

)( × mnnmx

1

)( × nllnx

1

)(

= lm

ml

x

· mn

nm

x

· nl

ln

x

= nl

ln

mn

nm

lm

ml

x

= lmn

lmmnlnlmmnln

x

= lmnx

0

= x0 = 1

22. Show that:

nm

nm

ab

ab

ba

ba

11

11

=

nm

b

a

Sol. nm

nm

ab

ab

ba

ba

11

11

=

nm

b

a

LHS = nm

nm

ab

ab

ba

ba

11

11

= nm

nm

a

ab

a

ab

b

ab

b

ab

11

11

=

n

n

m

m

nm

nm

a

ab

a

ab

bb

abab

)1()1(

·

)1()1(

= nmnm

nmnm

ababbb

aaabab

)1()1(·

)1()1(

= nm

nm

bb

ca

.

. = nm

nm

b

a

=

nm

b

a

= RHS

23. (i) If a = xm + ny1, b = xn + l ym and c = xl + m

yn, prove that am – n bn – 1 cl – m = 1.

(ii) If x = am + n, y = an + 1 and z = al + m, prove

that xm yn zl = xn yl zm.

Sol. (i) a = xm + n·yl, b = xn + lym, c = xl + myn

am – n·bn – l·cl – m = 1

LHS = am – n·bn – l·cl – m



= (xm + n·yl)m – n·(xn + l·ym)n – l·(xl + m·yn)l – m

= x(m + n) (m – n)·yl(m – n) × x(n + l) (n – l)·ym(n – l)

× x(l + m) (l – m)·yn(l – m)

= 22

nmx · ylm – ln · 22

lnx · ymn – ml · 22

mlx

· ynl – nm

= 222222

mllnnmx · ylm – ln + mn – ml + nl – mn

= x0 · y0 = 1 × 1 = 1 = RHS

(ii) x = am + n · y = an + 1 · z = al + m,

xm yn zl = xn yl zm

LHS = xm · yn · zl = a(m + n)m · a(n + l)n · a(l + m)l

= mnma 2

·nlna 2

·nlla 2

= lmnlmnlnma 222

RHS = xn · yl · zm

= (am + n)n · (yn + l)l · (al + m)m

= 2

nmna ·

2lnla

·2

mlma

= 222 mlmlnlnmna

= nlmnlmnmla 222

LHS = RHS

VERY SHORT ANSWER TYPE QUESTIONS

(VSAQs)

1. Write (625)–1/4 in decimal form.

Sol. 4

1

)625(

= 4

1

4 )5(

= 4

14

5

= 5–1

= 5

1 = 0.2

2. State the product law of exponents:

Sol. xm × xn = xm + n

3. State the quotient law of exponents.

Sol. xm xn = xm – n

4. State the power law of exponents.

Sol. (xm)n = xm × n = xmn

5. If 24 × 42 = 16x, then find the value of x.

Sol. 24 × 42 = 16x

24 × (22)2 = [(2)4]x

24 × 24 = 24x

24 + 4 = 24x 28 = 24x

Comparing, we get

4x = 8 x = 4

8 = 2

x = 2

6. If 3x – 1 = 9 and 4y + 2 = 64, what is the value

of y

x?

Sol. 3x – 1 = 9 3x – 1 = 32

Comparing, we get

x – 1 = 2 x = 2 + 1 = 3

x = 3

and 4y + 2 = 64 (4)y + 2 = (4)3

Comparing, we get

y + 2 = 3 y = 3 – 2 = 1

y = 1

Now y

x =

1

3 = 3

7. Write the value of 3 7 × 3 49 .

Sol. 3 7 × 3 49 = 3 497 = 3

1

)777(

= 3

13

7

= 71 = 7

8. Write 2

1

9

1

× 3

1

)64(

as a rational number..

Sol.2

1

9

1

× 3

1

)64(

=

2

12

3

× 3

1

3)4(

=

2

12

3

1 ×

3

13

4

=

1

3

1

× 4–1



= 3 × 4

1 =

4

3

9. Write the value of 3 27125 .

Sol. 3 27125 = 3 33 35

= 31

3)35( = 3

13

)15(

= (15)1 = 15

10. For any positive real number x, find the

value of

ba

b

a

x

x

×

cb

c

b

x

x

×

ac

a

c

x

x

Sol.

ba

b

a

x

x

×

cb

c

b

x

x

×

ac

a

c

x

x

= (xa – b)a + b.(xb – c)b + c.(xc – a)c + a

= x(a – b) (a + b).x(b – c) (b + c).x(c – a) (c + a)

= 22

bax .

22cbx

.22

acx

= 222222

accbbax = x0 = 1

11. Write the value of {5(81/3 + 271/3)3}1/4.

Sol.

4

1

3

3

1

3

1

2785

= 4

13

3

1

3

1

4

1

2785

= 4

3

3

1

3

1

4

1

2785

= 4

3

3

1

33

1

34

1

)3()2(5

= 4

34

1

325 = 4

1

5 × 4

3

5

= 4

3

4

1

5

= 51 = 5

12. Simplify :

2

4

1

2

1

)625(

.

Sol.

2

4

1

2

1

)625(

= 2

4

1

2

1

)625(

= 4

1

)625(

= 4

1

4 )5( = 4

14

5

= 51 = 5

13. For any positive real number x, write the

value of

abbax1

)( bccbx1

)( caacx1

)(

Sol. abbax1

)( . bccbx1

)( . caacx1

)(

= ababx1

. bcbcx1

. cacax1

= ab

ab

x. bc

bc

x. ca

ca

x

= x1 × x1 × x1 = x × x × x = x3

14. If (x – 1)3 = 8, what is the value of (x + 1)2?

Sol. (x – 1)3 = 8 (x – 1)3 = 23

Comparing, we get

x – 1 = 2 x = 2 + 1 = 3

x = 3

Now (x + 1)2 = (3 + 1)2 = (4)2 = 16

MULTIPLE CHOICE QUESTIONS (MCQs)


following:

1. The value of {2 – 3 (2 – 3)3}3 is

(a) 5 (b) 125



(c)5

1(d) –125

Sol. {2 – 3 (2 – 3)3}3 = {2 – 3 (–1)3}3

= {2 – 3 × (–1)}3

= (2 + 3)3 = (5)3

= 125 (b)

2. The value of x – yx – y when x = 2 and

y = –2 is

(a) 18 (b) –18

(c) 14 (d) –14

Sol. x = 2, y = –2

x – yx – y = 2 – (–2)2 – (–2)

= 2 – (–2)2 + 2 = 2 – (–2)4

= 2 – (+16) = 2 – 16 = –14 (d)

3. The product of the square root of x with

the cube root of x, is

(a) cube root of the square root of x

(b) sixth root of the fifth power of x

(c) fifth root of the sixth power of x

(d) sixth root of x

Sol. Product of sixth root of = 6 x × 3 x

= 6

1

x × 3

1

x = 3

1

2

1

x

= 6

23

x = 6

5

x = 6 5x (b)

4. The seventh root of x divided by the eighth

root of x is

(a) x (b) x

(c) 56 x (d) 56

1

x

Sol. Seventh root of x = 7 x = 7

1

x

Eighth root of x = 8 x = 8

1

x

8

1

7

1

x

x = 8

1

7

1

x = 56

78

x = 56

1

x = 56 x (c)

5. The square root of 64 divided by the cube

root of 64 is

(a) 64 (b) 2

(c)2

1(d) 3

2

64

Sol. 64 3 64

= 28 3 34 2

1

2 )8( 3

1

3)4(

= 2

12

8

3

13

4

= 8 4

= 4

8 = 2 (b)

6. Which of the following is (are) not equal to

6

1

5

1

6

5

?

(a)6

1

5

1

6

5

(b)

6

1

5

1

6

5

1

(c)30

1

5

6

(d)30

1

6

5

Sol.

6

1

5

1

6

5

=

6

1

5

1

6

5 =

30

1

6

5

(d)

7. When simplified (x–1 + y–1)–1 is equal to

(a) xy (b) x + y

(c)yx

xy

(d)xy

yx

Sol. (x–1 + y–1)–1 = x

1 +

y

1



= xy

xy =

xy

yx (d)

8. If 8x + 1 = 64, what is the value of 32x + 1?

(a) 1 (b) 3

(c) 9 (d) 27

Sol. 8x + 1 = 64 = 82

Comparing, we get

x + 1 = 2 x = 2 – 1 = 1

Now 32x + 1 = 32 × 1 + 1 = 32 + 1

= 33 = 3 × 3 × 3 = 27 (d)

9. If (23)2 = 4x, then 3x =

(a) 3 (b) 6

(c) 9 (d) 27

Sol. (23)2 = 4x 23 × 2 = (22)x

26 = 22x

Comparing,

2x = 6 x = 2

6 = 3

Now 3x = (3)3 = 3 × 3 × 3 = 27 (d)

10. If x–2 = 64, then 3

1

x + x0 =

(a) 2 (b) 3

(c)2

3(d)

3

2

Sol. x–2 = 64 2

1

x = 64

21

x = (8)2

x

1 = 8 x =

8

1

3

1

x + x0 = 3

1

8

1

+ 1

=

3

1

3

2

1

+ 1 = 3

13

2

1

+ 1

= 2

1 + 1 =

2

3(c)

11. When simplified 3

2

27

1

is

(a) 9 (b) –9

(c)9

1(d) –

9

1

Sol.3

2

27

1

=

3

2

3

3

1

= 3

23

3

1

=

2

3

1

= (–3)2 = (–3) × (–3) = 9 (a)

12. Which one of the following is not equal to

2

13 8

?

(a) 2

13 2

(b) 6

1

8

(c)

2

13 8

1(d)

2

1

Sol. 2

13 8

=

2

1

3

1

3)2(

=

2

1

3

13

2 = 2

1

2

=

2

1

(a) 2

13 2

=

2

1

3

1

)2(

=

2

1

3

1

2

= 6

1

2

(b) 6

1

8

= 6

1

3)2(

= 6

13

2

= 2

1

2

= 2

1



(c)

2

13 8

1 =

2

1

3

1

3 )2(

1

=

2

1

3

13

2

1

=

2

1

2

1 =

2

1

(d)2

1

We see that (a) is not equal (a)

13. Which one of the following is not equal to

2

3

9

100

?

(a)2

3

100

9

(b)

2

3

9

100

1

(c)10

3 ×

10

3 ×

10

3

(d)9

100

9

100

9

100

Sol.2

3

9

100

=

2

3

2

3

10

=

2

32

3

10 =

3

3

10

=

3

10

3

(a)2

3

100

9

= 2

32

10

3

=

3

10

3

(b)

2

3

9

100

1

=

2

32

3

10

1

= 3

3

10

1

=

3

10

3

(c)10

3 ×

10

3 ×

10

3 =

3

10

3

(d)9

100

9

100

9

100 =

3

9

100

= 2

3

9

100

= 2

32

3

10

=

3

3

10

= 9

100

9

100

9

100

(d)

14. If a, b, c are positive real numbers, then

ba 1 × cb 1 × ac 1 is equal to

(a) 1 (b) abc

(c) abc (d)abc

1

Sol. ba 1 × cb 1 × ac 1

= a

b ×

b

c ×

c

a

= c

a

b

c

a

b = 1 = 1 (a)

15. If

x

3

2x2

2

3

= 16

81, then x =

(a) 2 (b) 3

(c) 4 (d) 1

Sol.

x

3

2×

x2

2

3

= 16

81

x

2

3×

x2

2

3

= 4

4

2

3 =

4

2

3

xx

2

2

3=

4

2

3

x

2

3=

4

2

3

Comparing, we get

x = 4 (c)



16. The value of

2

1

23

4

28

is

(a)2

1(b) 2

(c)4

1(d) 4

Sol.

2

1

23

4

28

=

2

1

23

4

32)2(

=

2

1

23

43

22

=

2

1

23

43

22

= 2

1

2

4

2

2

= 2

1

4

2

2

2

= 2

1

242

1

= 2

1

22

1

=

2

12

2

1

=

2

1(a)

17. If a, b, c are positive real numbers, then

5 105103125 cba is equal to

(a) 5a2bc2 (b) 25ab2c

(c) 5a3bc3 (d) 125a2bc2

Sol. 5 105103125 cba = 5 1051055 cba

5 31255 6255 1255 255 5

1

= 5

11051055 cba

= 5

15

5

. 5

110

a . 5

15

b . 5

110

c

= 5a2bc2 (a)

18. If a, m, n are positive integers, then

mnm n a is equal to

(a) amn (b) a

(c) n

m

a (d) 1

Sol. mnm n a =

mn

m

na

1

1

=

mn

mna

1

= mn

mn

a = a (b)

19. If x = 2 and y = 4, then

yx

y

x

+

xy

x

y

=

(a) 4 (b) 8

(c) 12 (d) 2

Sol. x = 2, y = 4

yx

y

x

+

xy

x

y

=

42

8

4

+

24

2

4

=

2

2

1

+ (2)2

= 22 + 22 = 4 + 4 = 8 (b)

20. The value of m for which

4

1

3

12

27

1

= 7m, is



(a) –3

1(b)

4

1

(c) –3 (d) 2

Sol.

4

1

3

12

27

1

= 7m

4

1

3

122

7

= 7m

4

1

3

1)2(27

= 7m

4

1

3

14

7

= 7m

4

1

3

4

7 = 7m

3

1

7

= 7m

Comparing both sides,

m = –3

1(a)

21. The value of {(23 + 22)2/3 + (150 – 29)1/2}2,

is

(a) 196 (b) 289

(c) 324 (d) 400

Sol. {(23 + 22)2/3 + (150 – 29)1/2}2

= 22

1

3

2

])29150()423[(

= 22

1

3

2

])121()27[(

= 22

1

23

2

3 ])11()3[( = (9 + 11)2

= (20)2 = 400 (d)

22. (256)0.16 × (256)0.09

(a) 4 (b) 16

(c) 64 (d) 256.25

Sol. (256)0.16 × (256)0.09 = (256)0.16 + 0.09

= (256)0.25

= 100

25

8 )2( = 4

1

8 )2(

= 4

18

2

= 22 = 4 (a)

23. If 102y = 25, then 10–y equals

(a)5

1 (b)

50

1

(c)625

1(d)

5

1

Sol. 102y = 25 = 52

(10y)2 = (5)2 10y = 5

Now 10–y = y10

1 =

5

1(d)

24. If 9x + 2 = 240 + 9x, then x =

(a) 0.5 (b) 0.2

(c) 0.4 (d) 0.1

Sol. 9x + 2 = 240 + 9x 9x + 2 – 9x = 240

9x.92 – 9x = 240 9x(92 – 1) = 240

9x(81 – 1) = 240 9x × 80 = 240

9x = 80

40 = 3 (32)x = 31

32x = 31

2x = 1 x = 2

1 = 0.5 (a)

25. If x is a positive real number and x2 = 2,

then x3 =

(a) 2 (b) 2 2

(c) 3 2 (d) 4

Sol. x2 = 2 x = 2

x3 = ( 2 )3 = 2 × 2 × 2

= 2 × 2 = 2 2 (b)



26. If 5.1x

x = 8x–1 and x > 0, then x =

(a)4

2(b) 2 2

(c) 4 (d) 64

Sol. 5.1x

x = 8x–1

5.1x

x =

x

8 5.1

2

x

x = 8

x2 – 1.5 = 8 x0.5 = 8

2

1

x = 8

Squaring,

x = (8)2 = 64

x = 64 (d)

27. If 3

2

tg + 2

1

4

t , what is the value of g

when t = 64?

(a)2

31(b)

2

33

(c) 16 (d)16

257

Sol. 3

2

tg + 2

1

4

t

= 3

2

t + 4 ×

2

1

1

t

= 3

2

)64( + 4 ×

2

1

64

1

= 3

2

3)4( + 4 ×

2

1

2 )8(

1

= 3

3

2

4

+ 4 ×

2

12

8

1

= 42 +

8

4

= 16 + 2

1 =

2

33(b)

28. If 4x – 4x – 1 = 24, then (2x)x equals

(a) 5 5 (b) 5

(c) 25 5 (d) 125

Sol. 4x – 4x – 1 = 24 4x – 14

4x

= 24

4x

4

11 = 24 4x ×

4

3 = 24

4x = 3

424 = 32

(22)x = 32 = (2)5

22x = 25

Comparing, we get

2x = 5 x = 2

5

(2x)x = 2

5

2

52

= 2

5

)5(

= 55 = 55555

= 5 × 5 5 = 25 5 (c)

29. When simplifed )4( 2/3

256 is

(a) 8 (b)8

1

(c) 2 (d)2

1

Sol. )4( 2/3

256

= – 2/34

14

4

= – 2

3

)4(

= – 2

3

2 )2(



= –

2

32

)2( = –(2)–3

= – 3)2(

1 =

8

1

Now,

8

1

)256(

1 =

8

1

8 )2(

1

=

8

18

2

1

= 12

1 =

2

1(d)

30. If 225

382 x

= x5

53

, then x =

(a) 2 (b) 3

(c) 5 (d) 4

Sol.225

382 x

= x5

53

32x – 8 × 5x = 53 × 225

8

2

3

3x

× 5x = 5 × 5 × 5 × 5 × 5 × 3 × 3

32x × 5x = 38 × 32 × 55

(32)x × 5x = 310 × 55

9x × 5x = 35 × 35 × 55

(45)x = (45)5

Comparing, we get

x = 5 (c)

31. The value of 3

1

64

3

2

3

1

6464 , is

(a) 1 (b)3

1

(c) –3 (d) –2

Sol. 3

1

64

3

2

3

1

6464

= 3

1

3)4(

3

23

3

13

)4(4

=

3

13

4

3

23

3

13

44

= 4–1[41 – 42] = 4

1[4 – 16]

= 4

1(–12) = –3 (c)

32. If n5 = 125, then n 645 =

(a) 25 (b)125

1

(c) 625 (d)5

1

Sol. n5 = 125 25

n = 53

Comparing, we get

2

n = 3 n = 6

Now n 645 =

n

1

645

= 6

1

645 = 6

1

6 )2(5 = 6

16

25

= 52 = 25 (a)

33. If (16)2x + 3 = (64)x + 3, then 42x – 2 =

(a) 64 (b) 256

(c) 32 (d) 512

Sol. (16)2x + 3 = (64)x + 3 (24)2x + 3 = (26)x + 3

28x + 12 = 26x + 18

Comparing, we get

8x + 12 = 6x + 18 8x – 6x = 18 – 12

2x = 6 x = 2

6 = 3

Now 42x – 2 = 42 × (3) – 2 = 4+6 – 2 = 44

= 4 × 4 × 4 × 4 = 256 (b)

34. If 2–m × m2

1 =

4

1,



then 14

1

1

2

1

5

1)4(

m

m

is equal to

(a)2

1(b) 2

(c) 4 (d) –4

1

Sol. 2–m × m2

1 =

4

1

2–m – m =

2

2

1

2–2m = 2–2

Comparing, we get

–2m = –2 m = 2

2

= 1

Now, 14

1

1

2

1

5

1)4(

m

m

= 14

1

12

12 )5()2( mm

= 14

1

)1(2

12

52m

m

= 14

1{2m + 5+m}

= 14

1{21 + 51} =

14

1 × 7 =

2

1(a)

35. If mn

nm

2

2 = 16, n

p

3

3 = 81 and a = 10

1

2 , then

122

2

)(

pnm

pnm

a

a =

(a) 2 (b)4

1

(c) 9 (d)8

1

Sol.mn

nm

2

2 = 16, n

p

3

3 = 81 and a = 10

1

2 , then

122

2

)(

pnm

pnm

a

a

2m + n – n + m = 24 22m = 24

Comparing, we get

2m = 4 m = 2

4 = 2

Given, n

p

3

3 = 81

3p – n = (3)4

p – n = 4

122

2

)(

pnm

pnm

a

a = a2m + n – p × am – 2n + 2p

= a2m + n – p + m – 2n + 2p = a3m – n + p

_ a = 10

1

2 =

pnm

3

10

1

2

= 10

3

2

pnm

= 10

23

2

pn

= 10

6

2

pn

= 10

46

2

= 10

10

2 = 2 (a)

36. If x

x

2

25

3

6561813 = 37, then x =

(a) 3 (b) –3

(c)3

1(d) –

3

1

Sol. x

x

2

25

3

6561813 = 37

35x × 812 × 6561 = 37 × 32x



3 65613 21873 7293 2433 813 273 93 3

1

x

x

2

5

3

3 =

656181

32

7

= 824

7

3)3(

3

35x – 2x = 88

7

33

3

= 37 – 8 – 8 = 3–9

33x = 3–9

Comparing,

3x = –9 x = 3

9 = –3

x = –3 (b)

37. If 0 < y < x, which statement must be

true?

(a) x – y = yx

(b) x + x = x2

(c) x y = y x

(d) xy = x y

Sol. 0 < y < x

Only xy = x y is true (d)

38. If 10x = 64, what is the value of 1

210

x

?

(a) 18 (b) 42

(c) 80 (d) 81

Sol. 10x = 64 x10 = 64

2

1

)10( x = 8 210

x

= 8

Now, 1

210

x

= 210

x

× 101

= 8 × 10 = 80 (c)

39. 1

12

52513

565

nn

nn

is equal to

(a)3

5(b) –

3

5

(c)5

3(d) –

5

3

Sol. 1

12

52513

565

nn

nn

= )5213(5

)565(51

12

n

n

= 5213

5652

= 1013

3025

= 3

5(b)

40. If n2 = 1024, then

44

2

3

n

=

(a) 3 (b) 9

(c) 27 (d) 81

Sol. n2 = 1024 2

1

)2( n = 1024 = 210

2 10242 5122 2562 1282 642 322 162 82 42 2

1

22

n

= 210

Comparing,

2

n = 10 n = 2 × 10 = 20

Now,

44

2

3

n

=

44

202

3

= 32(5 – 4) = 32 × 1

= 32 = 9 (b)




1. Some Identies :

(i) 2a = a

(ii) a . b = ab

(iii)b

a =

b

a

(iv) ba ba = a – b

(v) ba ba = a2 – b

(vi) 2ba = a + b + 2 ab

(vii) ba dc = ac + ad +

bc + bd

2. Rationalisation Factors : Rationalisation

factor

(i) Of a is a

(ii) Of ba is ba

and of ba is ba

(iii) Of a

1 is a and if a is

a

1

EXERCISE 3.1

1. Simplify each of the following:

(i) 3 4 × 3 16 (ii) 4

4

2

1250

Sol. (i) 3 4 × 3 16 = 3 164

(_ a × b = ab )

= 3 64 = 3 444

= 3

1

3)4( = 3

13

4 = 41 = 4

(ii) 4

4

2

1250 = 4

2

1250

b

a

b

a

= 4 625

= 4 5555

= 4

1

4 )5( = 4

14

5

= 51 = 5

2. Simplify the following expressions:

(i) (4 + 7 ) (3 + 2 )

(ii) (3 + 3 ) (5 – 2 )

(iii) ( 5 – 2) ( 3 – 5 )

Sol. (i) (4 + 7 ) (3 + 2 )

= 4 × 3 + 4 2 + 3 7 + 7 × 2

= 12 + 4 2 + 3 7 + 14

(ii) (3 + 3 ) (5 – 2 )

= 3 × 5 – 3 2 + 5 3 – 3 × 2

= 15 – 3 2 + 5 3 – 6

(iii) ( 5 – 2) ( 3 – 5 )

= 5 × 3 – 5 × 5 – 2 3 + 2 5

= 15 – 25 – 2 3 + 2 5

= 15 – 5 – 2 3 + 2 5

= 15 – 2 3 + 2 5 – 5


(i) (11 + 11 ) (11 – 11 )

(ii) (5 + 7 ) (5 – 7 )

(iii) ( 8 – 2 ) ( 8 + 2 )

RATIONALISATION3



(iv) (3 + 3 ) (3 – 3 ) [NCERT]

(v) ( 5 – 2 ) ( 5 + 2 ) [NCERT]

Sol. (i) (11 + 11 ) (11 – 11 )

= (11)2 – ( 11 )2

{_ (a + b) (a – b) = a2 – b2}

= 121 – 11 = 110

(ii) (5 + 7 ) (5 – 7 )

= (5)2 – ( 7 )2

{_ (a + b) (a – b) = a2 – b2}

= 25 – 7 = 18

(iii) ( 8 – 2 ) ( 8 + 2 )

( 8 )2 – ( 2 )2

{_ (a + b) (a – b) = a2 – b2}

= 8 – 2 = 6

(iv) (3 + 3 ) (3 – 3 )

= (3)2 – ( 3 )2

{_ (a + b) (a – b) = a2 – b2}

= 9 – 3 = 6

(v) ( 5 – 2 ) ( 5 + 2 )

= ( 5 )2 – ( 2 )2

{_ (a + b) (a – b) = a2 – b2}

= 5 – 2 = 3


(i) ( 3 + 7 )2 (ii) ( 5 – 3 )2

(iii) (2 5 + 3 2 )2

Sol. ( 3 + 7 )2

= ( 3 )2 + ( 7 )2 + 2 × 3 × 7

{_ (a + b)2 = a2 + b2 + 2ab}

= 3 + 7 + 2 21 = 10 + 2 21

(ii) ( 5 – 3 )2

= ( 5 )2 + ( 3 )2 – 2 × 5 × 3

{_ (a – b)2 = a2 + b2 – 2ab}

= 5 + 3 – 2 15 = 8 – 2 15

(iii) (2 5 + 3 2 )2

= (2 5 )2 + (3 2 )2 + 2 × 2 5 × 3 2

{_ (a + b)2 = a2 + b2 + 2ab}

= 4 × 5 + 9 × 2 + 2 × 2 × 3 × 5 × 2

= 20 + 18 + 12 10

= 38 + 12 10

EXERCISE 3.2

1. Rationalise the denominators of each of the

following (i – vii):

(i)5

3(ii)

52

3

(iii)12

1(iv)

5

2

(v)2

13 (vi)

3

52

(vii)5

23

Sol. (i) 5

3 =

55

53

=

5

53 =

5

35

(ii)52

3 =

552

53

= 52

53

=

10

53 =

10

35

(iii)12

1 =

34

1

=

32

1



= 332

31

=

32

3

=

6

3

(iv)5

2 =

55

52

=

5

10 =

5

110

(v)2

13 =

22

2)13(

=

2

26

(vi)3

52 =

33

3)52(

= 3

156

(vii)5

23 =

55

523

=

5

103

= 5

310

2. Find the value to three places of decimals

of each of the following. It is given that

2 = 1.414, 3 = 1.732, 5 = 2.236

and 10 = 3.162.

(i)3

2(ii)

10

3

(iii)2

15 (iv)

2

1510

(v)3

32 (vi)

5

12

Sol. 2 = 1.414, 3 = 1.732, 5 = 2.236

and 10 = 3.162

(i)3

2 =

33

32

(Rationalising the denominator)

= 3

32 =

3

732.12 =

3

464.3 = 1.154

(ii)10

3 =

1010

103

=

10

103


= 10

)162.3(3 =

10

486.9 = 0.9486

(iii)2

15 =

22

2)15(


= 2

210 =

2

414.1162.3

= 2

576.4 = 2.288

(iv)2

1510 =

22

2)1510(


= 2

3020 =

2

31054

= 2

31052

= 2

732.1162.3)236.2(2

= 2

476.5472.4 =

2

948.9 = 4.974

(v)3

32 =

3

732.12 =

3

732.3 = 1.244

(vi)5

12 =

55

5)12(




= 5

510 =

5

236.2162.3

= 5

926.0 = 0.185

3. Express each one of the following with

rational denominator:

(i)23

1

(ii)

56

1

(iii)541

16

(iv)

5335

30

(v)352

1

(vi)

322

13

(vii)246

246

(viii)

352

123

(ix)aba

b

2

2

Sol. (i) 23

1

=

)23)(23(

)23(1

{Multiplying and dividing by (3 – 2 )}

= 22)2()3(

23

{_ (a + b) (a – b) = a2 – b2}

= 29

23

=

7

23

(ii)56

1

=

)56)(56(

)56(1

(Multiplying and dividing by 6 + 5 )

=

)5()6(

562

{_ (a + b) (a – b) = a2 – b2}

= 56

56

=

1

56

= 6 + 5

(iii)541

16

=

)541)(541(

)541(16

{Multiplying and dividing by ( 41 + 5)}

= 22)5()41(

)541(16

{_ (a + b) (a – b) = a2 – b2}

= 2541

)541(16

= 16

)541(16

= 41 + 5

(iv)5335

30

=

)5335)(5335(

)5335(30

{Multiplying and dividing by (5 3 +

3 5 )}

= 22 )53()35(

)5335(30

= 59325

)5335(30

= 4575

)5335(30

= 30

)5335(30 = 5 3 + 3 5

(v)352

1

=

)352)(352(

)352(1

{Multiplying and dividing by (2 5 + 3 )}

= 22)3()52(

352

=

320

352

= 17

352



(vi)322

13

=

)322)(322(

)322)(13(

{Multiplying and dividing by (2 2 + 3 )}

= 22)3()22(

32233223

= 38

322362

= 5

332262

(vii)246

246

=

)246)(246(

)246)(246(

{Multiplying and dividing by (6 – 4 2 )}

= 22

2

)24()6(

)246(

= 3236

246221636

= 4

2483236 =

4

24868

= 17 – 12 2

(viii)352

123

=

)352)(352(

)352)(123(

{Multiplying and dividing by (2 5 + 3)}

= 22)3()52(

3522335223

= 920

35229106

= 11

35229106

(ix)aba

b

2

2

= ))((

)(

2222

222

abaaba

abab

{Dividing and multiplying by ( 22 ba –a)}

= 2222

222

)(

)(

aba

abab

=

222

222)(

aba

abab

= 2

222)(

b

abab = 22 ba – a

4. Rationales the denominator and simplify:

(i)23

23

(ii)

347

325

(iii)223

21

(iv)

6253

562

(v)1848

2534

(vi)

3322

532

Sol. (i) 23

23

=

)23)(23(

)23)(23(


= 22

2

)2()3(

)23(

=

23

23223

= 1

625 = 5 – 2 6

(ii)347

325

=

)347)(347(

)347)(325(


= 22 )34()7(

33831432035



= 4849

243635

=

1

3611

= 11 – 6 3

(iii)223

21

=

)223)(223(

)223)(21(


= 22)22()3(

22223223

= 89

4253

= 1

257

= 7 + 5 2

(iv)6253

562

=

)6253)(6253(

)6253)(562(


= 22)62()53(

302553664306

= 6459

5364304

= 2445

3041524

= 21

3049

(v)1848

2534

=

)1848)(1848(

)1848)(2534(


= 22)18()48(

3659655441444

= 1848

656165694124

= 30

3064563448

= 30

6206123048 =

30

6818

= 15

649 (Dividing by 2)

(vi)3322

532

=

)3322)(3322(

)3322)(532(


= 22)33()22(

15310233664

= 278

1531021864

= 19

1536410218

= 19

1536410218

5. Simplify:

(i)35

35

+

35

35

(iii)32

1

+

35

2

+

52

1

(v)35

2

+

23

1

–

25

3

Sol. (i) 35

35

+

35

35

= )35)(35(

)35()35( 22

= 22

22

)3()5(

])3()5[(2

{_ (a + b)2 + (a – b)2 = 2(a2 + b2)}



= 35

)35(2

= 2

82 = 8

(iii)32

1

+

35

2

+

52

1

= 32

1

=

)32)(32(

)32(1

= 22 )3()2(

)32(1

= 34

32

= 1

32 = 2 – 3

= 35

2

=

)35)(35(

)35(2

= 22 )3()5(

)35(2

= 35

)35(2

=

2

)35(2 = 5 + 3

= 52

1

=

)52)(52(

)52(1

= 22)5()2(

52

=

54

52

= 1

52

= –(–2 – 5 )

Now, 32

1

+

35

2

+

52

1

= (2 – 3 ) + ( 5 + 3 ) + (–2 – 5 )

= 2 – 3 + 5 + 3 – 2 – 5 = 0

(v)35

2

+

23

1

–

25

3

= 35

2

=

)35)(35(

)35(2

= 22)3()5(

)35(2

=

35

)35(2

= 2

)35(2 = 5 – 3

= 23

1

=

)23)(23(

)23(1

= 22)2()3(

23

=

23

23

= 1

23 = 3 – 2

= 25

3

=

)25)(25(

)25(3

= 22 )2()5(

)25(3

=

25

)25(3

= 3

)25(3 = 5 – 2

Now, 35

2

+

23

1

–

25

3

= 5 – 3 + 3 – 2 – ( 5 – 2 )

= 5 – 3 + 3 – 2 – 5 + 2 = 0

6. In each of the following determine rational

numbers a and b:

(i)13

13

= a – b 3

(ii)22

24

= a – b

(iii)23

23

= a + b 2



(iv)347

335

= a + b 3

(v)711

711

= a – b 77

(vi)534

534

= a + b 5

Sol. (i) 13

13

= a – b 3

= 13

13

=

)13)(13(

)13)(13(


= 22

2

)1()3(

)13(

=

13

3213

= 2

324

= 2 – 3

Now, 2 – 3 = a – b 3

Comparing, we get

a = 2, b = 1

(ii)22

24

= a – b

= 22

24

=

)22)(22(

)22)(24(


= 22 )2()2(

222248

=

24

226

= 2

226 = 3 – 2

Now, 3 – 2 = a – b

Comparing, we get,

a = 3, b = 2

(iii)23

23

= a + b 2

= 23

23

=

)23)(23(

)23)(23(


= 22

2

)2()3(

)23(

=

29

23229

= 7

2611 =

7

11 +

7

62

Now, 7

11 +

7

62 = a + b 2

Comparing, we get

a = 7

11, b =

7

6

(iv)347

335

= a + b 3

= 347

335

=

)347)(347(

)347)(335(


= 22)34()7(

31232132035

= 4849

36335

=

1

13 = –1 + 3

Now, –1 + 3 = a + b 3

Comparing, we get

a = –1, b = 1

(v)711

711

= a – b 77

= 711

711

=

)711)(711(

)711)(711(




= 22

2

)7()11(

)711(

= 711

7112711

= 4

77218 =

2

779

2

9 –

2

177 = a – b 77

Comparing, we get

a = 2

9, b =

2

1

(vi)534

534

= a + b 5

= 534

534

=

)534)(534(

)534)(534(


= 22

2

)53()4(

)534(

=

4516

5244516

= 29

52461

29

52461

= a + b 5

29

61 –

29

245 = a + b 5

Comparing, we get

a = 29

61, b =

29

24

7. Find the value of 35

6

, it being given

that 3 = 1.732 and 5 = 2.236.

Sol. 3 = 1.732, 5 = 2.236

Now, 35

6

=

)35)(35(

)35(6


= 22)3()5(

)35(6

=

35

)35(6

= 2

)35(6 = 3( 5 + 3 )

= 3[2.236 + 1.732] = 3(3.968)

= 11.904

8. Find the values of each of the following

correct to three places of decimals, it being

given that 2 = 1.4142, 3 = 1.732,

5 = 2.2360, 6 = 2.4495 and 10 =

3.162.

(i)523

53

(ii)

223

21

Sol. 2 = 1.4142, 3 = 1.732, 5 = 2.2360,

6 = 2.4495 and 10 = 3.162

(i)523

53

=

)523)(523(

)523)(53(


= 22 )52()3(

5253569

=

209

59109

= 11

5919

=

11

2360.2919

= 11

124.2019

= 11

124.1

= 0.102

(ii)223

21

=

)223)(223(

)223)(21(




= 22)22()3(

2223223

=

89

2543

= 1

257 = 7 + 5(1.4142)

= 7 + 7.0710 = 14.071

9. Simplify:

(i)3223

3223

+

23

12

(ii)53

537

–

53

537

Sol. (i) 3223

3223

+

23

12

= 3223

3223

=

)3223)(3223(

)3223)(3223(


= 22

2

)32()23(

)3223(

= 3429

322323429

= 1218

6121218

=

6

61230

= 5 – 2 6

and 23

12

=

)23)(23(

)23)(34(

= 22)2()3(

)23(32

=

23

6232

= 1

626 = 6 + 2 6

3223

3223

+

23

12

= 5 – 2 6 + 6 + 2 6 = 111

(ii)53

537

–

53

537

= )53)(53(

)53)(537()53)(537(

= 22)5()3(

)553595721()53595721(

= 59

)155221()155221(

= 4

)526()526(

= 4

526526 =

4

54 = 5

10. If x = 2 + 3 , find the value of x3 + 3

1

x.

Sol. x = 2 + 3

x

1 =

32

1

=

)32)(32(

)32(1


= 22)3()2(

32

=

34

32

= 2 – 3

x + x

1 = 2 + 3 + 2 – 3 = 4

Cubing both sides,

31

xx = (4)3

x3 + 3

1

x + 3

xx

1 = 64

x3 + 3

1

x + 3 × 4 = 64



x3 + 3

1

x + 12 = 64

x3 + 3

1

x = 64 – 12 = 52

Hence x3 + 3

1

x = 52

11. If x = 3 + 8 , find the value of x2 + 2

1

x.

Sol. x = 3 + 8

x

1 =

83

1

=

)83)(83(

)83(1


= 22 )8()3(

83

=

89

83

=

1

83

= 3 – 8

Now, x + x

1 = 3 + 8 + 3 – 8 = 6

Squaring both sides

21

xx = (6)2

x2 + 2

1

x + 2 = 36

x2 + 2

1

x = 36 – 2 = 34

12. If x = 2

13 , find the value of 4x3 + 2x2 –

8x + 7.

Sol. x = 2

13 2x = 3 + 1

2x – 1 = 3


(2x – 1)2 = ( 3 )2 4x2 – 4x + 1 = 3

4x2 – 4x + 1 – 3 = 0 4x2 – 4x – 2 = 0

2x2 – 2x – 1 = 0

Now, 4x3 + 2x2 – 8x + 7

= (4x3 – 4x2 – 2x) + (6x2 – 6x – 3) + 10

= 2x(2x2 – 2x – 1) + 3(2x2 – 2x – 1) + 10

= 2x × 0 + 3 × 0 + 10

= 0 + 0 + 10 = 10


(VSAQs)

1. Write the value of (2 + 3 ) (2 – 3 ).

Sol. (2 + 3 ) (2 – 3 ) = (2)2 – ( 3 )2

{_ (a + b) (a – b) = a2 – b2}

= 4 – 3 = 1

2. Write the reciprocal of 5 + 2 .

Sol. Reciprocal of 5 + 2 = 25

1

(Rationalising its denominator)

= )25)(25(

)25(1

= 22 )2()5(

25

=

225

25

=

23

25

3. Write the rationalisation factor of 7 – 3 5 .

Sol. Rationalising factor of 7 – 3 5 is 7 +

3 5

{_ ( a + b ) ( a – b ) = a – b}

4. If 13

13

= x + y 3 , find the value of x

and y.

Sol.13

13

=

)13)(13(

)13)(13(




= 22

2

)1()3(

)13(

=

13

3213

= 2

324

= 2 – 3

2 – 3 = x + y 3

Comparing, we get

x = 2, y = –1

5. If x = 2 – 1, then write the value of x

1.

Sol. x = 2 – 1

x

1 =

12

1

=

)12)(12(

)12(1


= 22)1()2(

12

=

12

12

= 1

12 = 2 + 1

6. If a = 2 + 1, then find the value of

a – a

1.

Sol. a = 2 + 1

Then, a

1 =

12

1

=

)12)(12(

)12(1


= 22 )1()2(

12

=

12

12

= 2 – 1

a – a

1 = ( 2 + 1) – ( 2 – 1)

= 2 + 1 – 2 + 1 = 2

7. If x = 2 + 3 , find the value of x + x

1.

Sol. x = 2 + 3

Then, x

1 =

32

1

=

)32)(32(

)32(1

= 22)3()2(

32

=

34

32

= 1

32

x + x

1 = 2 + 3 + 2 – 3 = 4

8. Write the rationalisation factor of 5 – 2.

Sol. Rationalisation factor of 5 – 2 is 5 + 2

as ( a + b ) ( a – b ) = a – b

9. Simplify : 223 .

Sol. 223

= 2212

= 122)1()2( 22

{Converting into (a + b)2 = a2 + b2 + 2ab}

= 2)12( = 2 + 1

10. Simplify : 223 .

Sol. 223

= 2212

{Converting into (a – b)2 = a2 + b2 – 2ab}

= 122)1()2( 22

= 2)12( = 2 – 1

11. If x = 3 + 2 2 , then find the value of x

– x

1.

Sol. x = 3 + 2 2



Then, x

1 =

223

1

=

)223)(223(

)223(1

= 22 )22()3(

223

=

89

223

=

1

223

x + x

1 = 3 + 2 2 + 3 – 2 2 = 6

Now,

21

xx = x +

x

1 – 2

= 6 – 2 = 4 = (2)2

x – x

1 = 2



following:

1. 10 × 15 is equal to

(a) 5 6 (b) 6 5

(c) 30 (d) 25

Sol. 10 × 15 = 1510

= 3552 = 5 6 (a)

2. 5 6 × 5 6 is equal to

(a) 5 36 (b) 5 06

(c) 5 6 (d) 5 12

Sol. 5 6 × 5 6 = 5 66 = 5 36 (a)

3. The rationalisation factor of 3 is

(a) – 3 (b)3

1

(c) 2 3 (d) –2 3

Sol. Rationalisation factor of 3 is 3

1

1

1

aa (b)

4. The rationalisation factor of 2 + 3 is

(a) 2 – 3 (b) 2 + 3

(c) 2 – 3 (d) 3 – 2

Sol. Rationalisation factor of 2 + 3 is 2 – 3

{_ ( a + b ) ( a – b ) = a – b} (a)

5. If x = 5 + 2, then x – x

1 equals

(a) 2 5 (b) 4

(c) 2 (d) 5

Sol. x = 5 + 2

x

1 =

25

1

=

)25)(25(

)25(1

= 22 )2()5(

25

=

45

25

=

1

25

= 5 – 2

x – x

1 = ( 5 + 2) – ( 5 – 2)

= 5 + 2 – 5 + 2 = 4 (b)

6. If 13

13

= a – b 3 , then

(a) a = 2, b = 1 (b) a = 2, b = –1

(c) a = –2, b = 1 (d) a = b = 1

Sol.13

13

= a – b 3

13

13

=

)13)(13(

)13)(13(




= 22

2

)1()3(

)13(

=

13

3213

= 2

324

= 2 – 3

Now, 2 – 3 = a – b 3

Comparing, we get

a = 2, b = 1 (a)

7. The simplest rationalising factor of 3 500

is

(a) 3 2 (b) 3 5

(c) 3 (d) none of these

Sol. 3 500 = 3 55522

Making triplet, we find 2 is required to

complete the triplet of 2

Required factor = 3 2 (a)

8. The simplest rationalising factor of 3 +

5 is

(a) 3 – 5 (b) 3 – 5

(c) 3 – 5 (d) 3 + 5

Sol. The simplest rationalising factor of

3 + 5 is 3 – 5 as ( a + b )

( a – b ) = a – b (c)

9. The simplest rationalising factor of 2 5 –

3 is

(a) 2 5 + 3 (b) 2 5 + 3

(c) 5 + 3 (d) 5 – 3

Sol. The simplest rationalising factor if

2 5 – 3 is 2 5 + 3 as ( a + b )

( a – b ) = a – b (b)

10. If x = 73

2

, then (x – 3)2 =

(a) 1 (b) 3

(c) 6 (d) 7

Sol. x = 73

2

=

)73)(73(

)73(2


= 22)7()3(

)73(2

=

79

)73(2

= 2

)73(2 = 3 – 7

Now, x – 3 = 3 – 7 – 3 = – 7

(x – 3)2 = (– 7 )2 = 7 (d)

11. If x = 7 + 4 3 and xy = 1, then 2

1

x +

2

1

y =

(a) 64 (b) 134

(c) 194 (d)49

1

Sol. x = 7 + 4 3 and xy = 1

y = x

1 =

347

1

y = x

1 =

)347)(347(

)347(1


= 22)34()7(

347

=

4849

347

= 1

347

= 7 – 4 3

Now, 2

1

x + 2

1

y



_ y = x

1

y

1 = x

2

1

x + 2

1

y = 2

1

x + x2

= x2 + 2

1

x =

21

xx – 2

= [(7 + 4 3 ) + 7 – 4 3 ]2 – 2

= (14)2 – 2 = 196 – 2 = 194 (c)

12. If x + 15 = 4, then x + x

1 =

(a) 2 (b) 4

(c) 8 (d) 1

Sol. x + 15 = 4 x = 4 – 15

and x

1 =

154

1

=

)154)(154(

)154(1


= 22)15()4(

154

=

1516

154

= 1

154 = 4 + 15

Now, x + x

1 = 4 – 15 + 4 + 15

= 8 (c)

13. If x = 35

35

and y =

35

35

, then x

+ y + xy =

(a) 9 (b) 5

(c) 17 (d) 7

Sol. x = 35

35

=

)35)(35(

)35)(35(


= 22

2

)3()5(

)35(

=

35

35235

= 2

1528 = 4 + 15

Similarly,

y = 35

35

=

)35)(35(

)35)(35(

= 22

2

)3()5(

)35(

=

35

35235

= 2

1528 = 4 – 15

Now, x + y + xy = 4 + 15 + 4 – 15 +

(4 + 15 ) (4 – 15 )

= 8 + [(4)2 – ( 15 )2] = 8 + (16 – 15)

= 8 + 1 = 9 (a)

14. If x = 23

23

and y =

23

23

, then x2

+ xy + y2 =

(a) 101 (b) 99

(c) 98 (d) 102

Sol. x = 23

23

=

)23)(23(

)23)(23(


= 22

2

)2()3(

)23(

=

23

23223

= 1

625 = 5 – 2 6

Similarly,

y = 23

23

=

)23)(23(

)23)(23(




= 22

2

)2()3(

)23(

=

23

23223

= 1

625 = 5 + 2 6

x2 + y2 + xy = (5 – 2 6 )2 + (5 + 2 6 )2 +

(5 – 2 6 ) (5 + 2 6 )

= 25 + 24 – 20 6 + 25 + 24 + 20 6 +

(5)2 – (2 6 )2

= 49 + 49 + 25 – 24 = 99 (b)

15.89

1

is equal to

(a) 3 + 2 2 (b)223

1

(c) 3 – 2 2 (d)2

3 – 2

Sol.89

1

=

243

1

=

223

1

= )223)(223(

)223(1

= 22 )22()3(

223

= 89

223

= 1

223 = 3 + 2 2 (a)

16. The value of 1827

3248

is

(a)3

4(b) 4

(c) 3 (d)4

3

Sol.1827

3248

=

2939

216316

= 2333

2434

=

)23(3

)23(4

=

3

4(a)

17. If 32

35

= x + y 3 , then

(a) x = 13, y = –7 (b) x = –13, y = 7

(c) x = –13, y = –7 (d) x = 13, y = 7

Sol.32

35

= x + y 3

32

35

=

)32)(32(

)32)(35(


= 22)3()2(

3323510

=

34

3713

= 1

3713 = 13 – 7 3

13 – 7 3 = x + y 3

Comparing, we get

x = 13, y = –7 (a)

18. If x = 3 32 , then x3 + 3

1

x =

(a) 2 (b) 4

(c) 8 (d) 9

Sol. x = 31

32 x3 = 2 + 3

3

1

x =

32

1

=

)32)(32(

)32(1

= 22 )3()2(

32

=

34

32


= 1

32 = 2 – 3



x3 + 3

1

x = 2 + 3 + 2 – 3 = 4 (b)

19. The value of 223 is

(a) 2 – 1 (b) 2 + 1

(c) 3 – 2 (d) 3 + 2

Sol. 223 = 2212

= 122)1()2( 22

= 2)12( = 2 – 1 (a)

20. The value of 625 is

(a) 3 – 2 (b) 3 + 2

(c) 5 + 6 (d) none of these

Sol. 625 = 23223

= 232)2()3( 22

= 2)23( = 3 + 2 (b)

21. If 2 = 1.4142, then 12

12

is equal to

(a) 0.1718 (b) 5.8282

(c) 0.4142 (d) 2.4142

Sol. 2 = 1.4142

12

12

=

)12)(12(

)12)(12(


= 22

2

)1()2(

)12(

=

12

)12( 2

= 1

)12( 2 = 2 – 1 = 1.4142 – 1

= 0.4142 (c)

22. If 2 = 1.414, then the value of 6 – 3

upto three places of decimal is

(a) 0.235 (b) 0.707

(c) 1.414 (d) 0.471

Sol. 2 = 1.414

6 – 3 = 32 – 3

= 3 × 2 – 3

= 3 ( 2 – 1)

= 1.732 (1.414 – 1)

= 1.732 × 0.414 = 0.717 (b)

23. The positive square root of 7 + 48 is

(a) 7 + 2 3 (b) 7 + 3

(c) 2 + 3 (d) 3 + 2

Sol. 7 + 48 = 7 + 4 × 12

= 7 + 2 12

= 4 + 3 + 2 34

= ( 4 )2 + ( 3 )2 + 2 × ( 4 × 3 )

= ( 4 + 3 )2

Square root = 4 + 3

= 2 + 3 (c)

24. If x = 6 + 5 , then x2 + 2

1

x – 2 =

(a) 2 6 (b) 2 5

(c) 24 (d) 20

Sol. x = 6 + 5

Then x

1=

56

1

=

)56)(56(

56

= 22)5()6(

56

=

56

56



= 1

56 = 6 – 5

Now, x2 + 2

1

x – 2 =

21

xx

= [( 6 + 5 ) – ( 6 – 5 )]2

= ( 6 + 5 – 6 + 5 )2 = (2 5 )2

= 4 × 5 = 20 (d)

25. If 1013 a = 8 + 5 , then a =

(a) –5 (b) –6

(c) –4 (d) –2

Sol. 1013 a = 8 + 5

1058 a = 8 + 5


( 1058 a )2 = ( 8 + 5 )2

8 + 5 – a 10 = 8 + 5 + 2 × 8 × 5

–a 10 = 2 40 = 2 × 104

–a 10 = 2 × 2 × 10

–a 10 = 4 10

Comparing, we get

–a = 4 a = –4 (c)




1. (a + b)2 = a2 + 2ab + b2

2. (a – b)2 = a2 – 2ab + b2

3. (a + b) (a – b) = a2 – b2

4. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc +

2ca

= a2 + b2 + c2 + 2(ab + bc + ca)

5. (x + a) (x + b) = x2 + (a + b)x + ab

6. (a + b)3 = a3 + b3 + 3ab(a + b)

= a3 + 3a2b + 3ab2 + b3

7. (a – b)3 = a3 – b3 – 3ab (a – b)

= a3 – 3a2b + 3ab2 – b3

8. a3 + b3 = (a + b)3 – 3ab(a + b)

= (a + b) (a2 – ab + b2)

9. a3 – b3 = (a – b)3 + 3ab(a – b)

= (a – b) (a2 + ab + b2)

10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 +

c2 – ab – bc – ca)

11. a3 + b3 + c3 = 3abc if (a + b + c) = 0

12. (a + b)3 – (a – b)3 = 2(b3 + 3a2b)

13. (a + b)3 + (a – b)3 = 2(b3 + 3ab2)

EXERCISE 4.1

1. Evaluate each of the following using

identities:

(i)

21

2

x

x (ii) (2x + y) (2x – y)

(iii) (a2b – b2a)2 (iv) (a – 0.1) (a + 0.1)

(v) (1.5x2 – 0.3y2) (1.5x2 + 0.3y2)

Sol. (i)

21

2

x

x = (2x)2 +

21

x– 2 × 2x ×

x

1

{_ (a – b)2 = a2 + b2 – 2ab}

= 4x2 + 2

1

x – 4 = 4x2 – 4 + 2

1

x

(ii) (2x + y) (2x – y)

= (2x + y) (2x – y)

= (2x)2 – (y)2

{_ (a + b) (a – b) = a2 – b2)

= 4x2 – y2

(iii) (a2b – b2a)2 = (a2b)2 + (b2a)2 – 2 × a2b ×

b2a

= a4b2 + b4a2 – 2a3b3

{_ (a – b)2 = a2 + b2 – 2ab}

= a4b2 – 2a3b3 + b4a2

(iv) (a – 0.1) (a + 0.1) = (a)2 – (0.1)2

{_ (a + b) (a – b) = a2 – b2}

= a2 – 0.01

(v) (1.5x2 – 0.3y2) (1.5x2 + 0.3y2)

{_ (a + b) (a – b) = a2 – b2}

= (1.5x2)2 – (0.3y2)2

= 2.25x4 – 0.09y4

2. Evaluate each of the following using

identities:

(i) (399)2 (ii) (0.98)2

(iii) 991 × 1009 (iv) 117 × 83

Sol. (i) (399)2 = (400 – 1)2

= (400)2 – 2 × 400 × 1 + (1)2

{_ (a – b)2 = a2 – 2ab + b2}

= 160000 – 800 + 1

= 160001 – 800 = 159201

(ii) (0.98)2 = (1 – 0.02)2

{_ (a – b)2 = a2 – 2ab + b2}

= (1)2 – 2 × 1 × 0.02 + (0.02)2

= 1 – 0.04 + 0.0004

= 1.0004 – 0.04 = 1.0004 – 0.0400

= 0.9604

(iii) 991 × 1009 = (1000 – 9) (1000 + 9)

{_ (a + b) (a – b) = a2 – b2}

= (1000)2 – (9)2

= 1000000 – 81 = 999919

(iv) 117 × 83 = (100 + 17) (100 – 17)

{_ (a + b) (a – b) = a2 – b2}

= (100)2 – (17)2

= 10000 – 289 = 9711


(i) 175 × 175 + 2 × 175 × 25 + 25 × 25

(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22

ALGEBRAIC IDENTITIES4



(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 ×

0.24

(iv)66.6

17.117.183.783.7

Sol. (i) 175 × 175 + 2 × 175 × 25 + 25 × 25

Let a = 175 and b = 25, then

a2 + 2ab + b2 = (a + b)2

= (175 + 25)2 = (200)2 = 40000

(ii) 322 × 322 – 2 × 322 × 22 + 22 × 22

Let a = 322 and b = 22, then

a2 – 2ab + b2 = (a – b)2

= (322 – 22)2 = (300)2 = 90000

(iii) 0.76 × 0.76 + 2 × 0.76 × 0.24 + 0.24 ×

0.24

Let a = 0.76 and b = 0.24, then

a2 + 2ab + b2 = (a + b)2

= (0.76 + 0.24)2 = (1.00)2 = 1

(iv)66.6

17.117.183.783.7

Let a = 7.83, b = 1.17, a – b = 7.83 – 1.17

= 6.66

ba

bbaa

=

ba

ba

22

= ba

baba

)()( = a + b

= 7.83 + 1.17 = 9.00 = 9

4. If x + x

1 = 11, find the value of x2 + 2

1

x.

Sol. x + x

1 = 111


21

xx = (11)2

x2 + 2

1

x + 2 = 121

x2 + 2

1

x = 121 – 2 = 119

x2 + 2

1

x = 119

5. If x – x

1 = –1, find the value of x2 + 2

1

x.

Sol. x – x

1 = –1


21

xx = (–1)2 x2 + 2

1

x – 2 = 1

x2 + 2

1

x = 1 + 2 = 3

6. If x + x

1 = 5 , find the values of x2 +

2

1

x and x4 + 4

1

x.

Sol. x + x

1 = 5


2

1

xx = ( 5 )2 x2 + 2

1

x + 2 = 5

(i) x2 + 2

1

x = 5 – 2 = 3

Again squaring,

2

2

2 1

x

x = (3)2

x4 + 4

1

x + 2 = 9

x4 + 4

1

x = 9 – 2 = 7

x2 + 2

1

x = 3, x4 + 4

1

x = 7

7. If 9x2 + 25y2 = 181 and xy = –6, find the

value of 3x + 5y.



Sol. 9x2 + 25y2 = 181, and xy = –6

(3x + 5y)2 = (3x)2 + (5y)2 + 2 × 3x + 5y

9x2 + 25y2 + 30xy

= 181 + 30 × (–6)

= 181 – 180 = 1

= (+1)2

3x + 5y = +1

8. If 2x + 3y = 8 and xy = 2, find the value of

4x2 + 9y2.

Sol. 2x + 3y = 8 and xy = 2

Now, (2x + 3y)2 = (2x)2 + (3y)2 + 2 × 2x × 3y

(8)2 = 4x2 + 9y2 + 12xy

64 = 4x2 + 9y2 + 12 × 2

64 = 4x2 + 9y2 + 24

4x2 + 9y2 = 64 – 24 = 40

4x2 + 9y2 = 40

9. If 3x – 7y = 10 and xy = –1, find the value

of 9x2 + 49y2

Sol. 3x – 7y = 10, xy = –1

3x – 7y = 10


(3x – 7y)2 = (10)2

(3x)2 + (7y)2 – 2 × 3x × 7y = 100

9x2 + 49y2 – 42xy = 100

9x2 + 49y2 – 42(–1) = 100

9x2 + 49y2 + 42 = 100

9x2 + 49y2 = 100 – 42 = 58

10. Simplify each of the following products:

(i)

ba 32

1

ab2

13

22 94

1ba

(ii)

3

7

nm

7

nm

Sol. (i)

ba 32

1

ab2

13

22 94

1ba

=

ba 32

1

ba 32

1

22 94

1ba

=

2

2

)3(2

1ba

22 94

1ba

{_ (a + b) (a – b) = a2 – b2}

=

22 94

1ba

22 94

1ba

=

2

2

4

1

a – (9b2)2

= 16

1a4 – 81b4

(ii)

3

7

nm

7

nm

=

2

7

nm

7

nm

7

nm

=

2

7

nm

2

2

7

nm

{_ (a + b) (a – b) = a2 – b2}

=

2

7

nm

49

22 n

m

11. If x2 + 2

1

x = 66, find the value of x –

x

1.

Sol. x2 + 2

1

x = 66

21

xx = x2 + 2

1

x – 2

= 66 – 2 = 64 = (+8)2

x – x

1 = +8

12. If x2 + 2

1

x = 79, find the value of x +

x

1.



Sol. x2 + 2

1

x = 79

21

xx = x2 + 2

1

x + 2

= 79 + 2 = 81 = (+9)2

x + x

1 = +9

13. Simplify each of the following products:

(i)

5

2

2

x

25

2 x – x2 + 2x

(ii) (x2 + x – 2) (x2 – x + 2)

(iii) (x3 – 3x2 – x) (x2 – 3x + 1)

(iv) (2x4 – 4x2 + 1) (2x4 – 4x2 – 1)

Sol. (i)

5

2

2

x

25

2 x – x2 + 2x

=

5

2

2

x

5

2

2

x – x2 + 2x

= –

5

2

2

x

5

2

2

x – x2 + 2x

= –

2

5

2

2

x

– x2 + 2x

= –

5

2

22

5

2

2

22xx

– x2 + 2x

= –

5

2

25

4

4

2 xx – x2 + 2x

= 4

2x

– 25

4 +

5

2x – x2 + 2x

= 4

2x

– x2 + 5

2x + 2x –

25

4

= –

4

4 22 xx +

5

102 xx –

25

4

= –4

52

x +

5

12x –

25

4

(ii) (x2 + x – 2) (x2 – x + 2)

= [x2 + (x – 2)] [x2 – (x – 2)]

= (x2)2 – (x – 2)2

{_ (a + b) (a – b) = a2 – b2}

= x4 – [(x)2 – 2 × x × 2 + (2)2]

= x4 – (x2 – 4x + 4)

= x4 – x2 + 4x – 4

(iii) (x3 – 3x2 – x) (x2 – 3x + 1)

= x(x2 – 3x – 1) (x2 – 3x + 1)

= x[(x2 – 3x) – 1] [(x2 – 3x) + 1]

= x[(x2 – 3x)2 – (1)2]

{_ (a + b) (a – b) = a2 – b2}

= x[(x2)2 – 2 × x2 × 3x + (3x)2 – 1]

= x[x4 – 6x3 + 9x2 – 1]

= x5 – 6x4 + 9x3 – x

(iv) (2x4 – 4x2 + 1) (2x4 – 4x2 – 1)

= [(2x4 – 4x2) + 1] [(2x4 – 4x2) – 1]

= (2x4 – 4x2)2 – (1)2

{_ (a + b) (a – b) = a2 + b2}

= (2x4)2 – 2 × 2x4 × 4x2 + (4x2)2 – 1

= 4x8 – 16x6 + 16x4 – 1

14. Prove that a2 + b2 + c2 – ab – bc – ca is

always non-negative for all values of a, b

and c.

Sol. a2 + b2 + c2 – ab – bc – ca

{Multiplying and dividing by 2}

= 2

1[2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca]

= 2

1[a2 + b2 – 2ab + b2 + c2 – 2bc + c2 +

a2 – 2ca]

= 2

1[(a – b)2 + (b – c)2 + (c – a)2]

_ The given expression is sum of these

squares

Its value is always positive

Hence the given expression is always non-

negative for all values of a, b and c



EXERCISE 4.2

1. Write the following in the expanded form:

(i) (a + 2b + c)2 (ii) (2a – 3b – c)2

(iii) (–3x + y + z)2 (iv) (m + 2n – 5p)2

(v) (2 + x – 2y)2 (vi) (a2 + b2 + c2)2

(vii) (ab + bc + ca)2 (viii)

2

x

z

z

y

y

x

(ix)

2

ab

c

ca

b

bc

a

(x) (x + 2y + 4z)2 [NCERT]

(xi) (2x – y + z)2 [NCERT]

(xii) (–2x + 3y + 2z)2 [NCERT]

Sol. We know that (a + b + c)2 = a2 + b2 + c2 +

2ab + 2bc + 2ca, Therefore

(i) (a + 2b + c)2 = (a)2 + (2b)2 + (c)2 + 2 × a

× 2b + 2 × 2b × c + 2 × c × a

= a2 + 4b2 + c2 + 4ab + 4bc + 2ca

(ii) (2a – 3b – c)2 = (2a)2 + (3b)2 + (c)2 + 2 ×

2a × –3b + 2 × –3b × –c + 2 × –c × 2a

= 4a2 + 9b2 + c2 – 12ab + 6bc – 4ca

(iii) (–3x + y + z)2 = (–3x)2 + (y)2 + z2 + 2 ×

(–3x) × y + 2 × y × z + 2 × z × (–3x)

= 9x2 + y2 + z2 – 6xy + yz – 6zx

(iv) (m + 2n – 5p)2 = (m)2 + (2n)2 + (–5p)2 + 2 ×

m × 2n + 2 × (2n) × (–5p) + 2 × (–5p) × m

= m2 + 4n2 + 25p2 + 4mn – 20np – 10pm

(v) (2 + x – 2y)2 = (2)2 + (x)2 + (–2y2) + 2 × 2

× x + 2 × x × (–2y) + 2 × (–2y) × 2

= 4 + x2 + 4y2 + 4x – 4xy – 8y

(vi) (a2 + b2 + c2)2 = (a2)2 + (b2)2 + (c2)2 + 2 ×

a2 × b2 + 2b2 × c2 + 2c2 × a2

= a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2

(vii) (ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 +

2ab × bc + 2bc × ca + 2ca × ab

= a2b2 + b2c2 + c2a2 + 2ab2c + 2bc2a + 2ca2b

(viii)

2

x

z

z

y

y

x =

2

y

x +

2

z

y +

2

x

z +

2 × y

x ×

z

y + 2 ×

z

y +

x

z + 2 ×

x

z ×

y

x

= 2

2

y

x + 2

2

z

y +

x

z2

+ 2z

x + 2

x

y + 2

y

z

(ix)

2

ab

c

ca

b

bc

a =

2

bc

a +

2

ca

b +

2

ab

c + 2 ×

bc

a ×

ca

b + 2

ca

b ×

ab

c +

2ab

c ×

bc

a

= 22

2

cb

a + 22

2

ac

b + 22

2

ba

c + 2

2

c + 2

2

a + 2

2

b

= 22

2

cb

a + 22

2

ac

b + 22

2

ba

c + 2

2

a + 2

2

b + 2

2

c

(x) (x + 2y + 4z)2 = (x)2 + (2y)2 + (4z)2 + 2 × x

× 2y + 2 × 2y × 4z + 2 × 4z × x

= x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx

(xi) (2x – y + z)2 = (2x)2 + (–y)2 + (z)2 + 2 ×

(2x) × (–y) + 2 × (–y) × z + 2z × 2x

= 4x2 + y2 + z2 – 4xy – 2yz + 4zx

(xii) (–2x + 3y + 2z)2 = (–2x)2 + (3y)2 + (2z)2 + 2

× (–2x) × 3y + 2 × 3y × 2z + 2 × 2z × (–2x)

= 4x2 + 9y2 + 4z2 – 12xy + 12yz – 8zx

2. If a + b + c = 0 and a2 + b2 + c2 = 16, find

the value of ab + bc + ca.

Sol. a + b + c = 0


(a + b + c)2 = 0 a2 + b2 + c2 + 2(ab + bc

+ ca) = 0

16 + 2(ab + bc + c) = 0

2(ab + bc + ca) = –16

ab + bc + ca = 2

16 = –8

ab + bc + ca = –8

3. If a2 + b2 + c2 = 16 and ab + bc + ca = 10,

find the value of a + b + c.

Sol. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

= 16 + 2 × 10 = 16 + 20 = 36

= (+6)2



a + b + c = +6

4. If a + b + c = 9 and ab + bc + ca = 23,

find the value of a2 + b2 + c2.

Sol. (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc +

ca)

(9)2 = a2 + b2 + c2 + 2 × 23

81 = a2 + b2 + c2 + 46

a2 + b2 + c2 = 81 – 46 = 35

a2 + b2 + c2 = 35

5. Find the value of 4x2 + y2 + 25z2 + 4xy –

10yz – 20zx when x = 4, y = 3 and z = 2.

Sol. x = 4, y = 3, z = 2

4x2 + y2 + 25z2 + 4xy – 10yz – 20zx

= (2x)2 + (y)2 + (5z)2 + 2 × 2x × y – 2 × y

× 5z – 2 × 5z × 2x

= (2x + y – 5z)2

= (2 × 4 + 3 – 5 × 2)2 = (8 + 3 – 10)2

= (11 – 10)2 = (1)2 = 1

6. Simplify:

(i) (a + b + c)2 + (a – b + c)2

(ii) (a + b + c)2 – (a – b + c)2

(iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2

(iv) (2x + p – c)2 – (2x – p + c)2

(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2

Sol. (i) (a + b + c)2 + (a – b + c)2

= (a2 + b2 + c2 + 2ab + 2bc + 2ca) + (a2 +

b2 + c2 – 2ab – 2bc + 2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + b2

+ c2 – 2ab – 2bc + 2ca

= 2a2 + 2b2 + 2c2 + 4ca

= 2(a2 + b2 + c2 + 2ca)

(ii) (a + b + c)2 – (a – b + c)2

= (a2 + b2 + c2 + 2ab + 2bc + 2ca) – (a2 +

b2 + c2 – 2ab – 2bc + 2ca)

= a2 + b2 + c2 + 2ab + 2bc + 2ca – a2 – b2

– c2 + 2ab + 2bc – 2ca

= 4ab + 4bc = 4(ab + bc)

(iii) (a + b + c)2 + (a – b + c)2 + (a + b – c)2

= a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + b2

+ c2 – 2ab – 2bc + 2ca + a2 + b2 + c2 +

2ab – 2bc – 2ca

= 3a2 + 3b2 + 3c2 + 2ab – 2bc + 2ca

= 3(a2 + b2 + c2) + 2(ab – bc + ca)

(iv) (2x + p – c)2 – (2x – p + c)2

= [(2x)2 + (p)2 + (–c)2 + 2 × 2x × p – 2pc

– 2 × c × 2x] – [(2x)2 + (–p)2 + (c)2 – 2 ×

2x × p – 2pc + 2c × 2x]

= (4x2 + p2 + c2 + 4xp – 2pc – 4cx) – (4x2

+ p2 + c2 – 4xp – 2pc + 4cx)

= 4x2 + p2 + c2 + 4xp – 2pc – 4cx – 4x2 – p2

– c2 + 4xp + 2pc – 4cx

= 8xp – 8cx

= 8x(p – c)

(v) (x2 + y2 – z2)2 – (x2 – y2 + z2)2

= [(x2)2 + (y2)2 + (–z2)2 + 2x2y2 – 2y2z2 –

2z2x2] – [(x2)2 + (–y2)2 + (z2)2 – 2x2y2 – 2y2z2

+ 2z2x2]

= x4 + y4 + z4 + 2x2y2 – 2y2z2 – 2z2x2 – x4 –

y4 – z4 + 2x2y2 + 2y2z2 – 2z2x2

= 4x2y2 – 4z2x2

= 4x2(y2 – z2)

7. Simplify each of the following expressions:

(i) (x + y + z)2 +

2

32

zy

x –

2

432

zyx

(ii) (x + y – 2z)2 – x2 – y2 – 3z2 + 4xy

(iii) (x2 – x + 1)2 – (x2 + x + 1)2

Sol. (i) (x + y + z)2 +

2

32

zy

x

–

2

432

zyx

= (x2 + y2 + z2 + 2xy + 2yz + 2zx) +

4

22 y

x + 9

2z

+ 2

2 yx +

2

2 y ×

3

z

+

xz

3

2 –

4

2x +

9

2y

+ 16

2z

+ 2

2x

× 3

y +

3

2y ×

4

z +

4

2z ×

2

x



= x2 + y2 + z2 + 2xy + 2yz + 2zx + x2 + 4

2y

+ 9

2z

+ xy + 3

yz +

3

2zx –

4

2x

– 9

2y

–

16

2z

– 3

xy –

6

yz –

4

xz

= x2 + x2 – 4

2x

+ y2 + 4

2y

– 9

2y + z2 +

9

2z –

16

2z

+ 2xy + xy – 3

xy + 2yz +

3

yz –

6

yz + 2zx +

3

2zx –

4

zx

= 4

44 222 xxx +

36

4936222

yyy +

144

916144222

zzz +

3

36 xyxyxy +

6

212 yzyzyz +

12

3824 zxzxzx

= 4

72

x +

36

412

y +

14

1512

z +

3

8xy +

6

13yz

+ 12

29zx

(ii) (x + y – 2z)2 – x2 – y2 – 3z2 + 4xy

= x2 + y2 + 4z2 + 2xy – 4yz – 4zx – x2 – y2 –

3z2 + 4xy

= z2 + 6xy – 4yz – 4zx

(iii) (x2 – x + 1)2 – (x2 + x + 1)2

= (x4 + x2 + 1 – 2x3 – 2x + 2x2) – (x4 + x2 +

1 + 2x3 + 2x + 2x2)

= x4 + x2 + 1 – 2x3 – 2x + 2x2 – x4 – x2 – 1

– 2x3 – 2x – 2x2

= –4x3 – 4x = –4x(x2 + 1)

EXERCISE 4.3

1. Find the cube of each of the following

binomial expressions:

(i)x

1 +

3

y(ii)

x

3 – 2

2

x

(iii) 2x + x

3(iv) 4 –

x3

1

Sol. (i)

3

3

1

y

x =

31

x +

3

3

y

+ 3

21

x ×

3

y + 3 ×

x

1 ×

2

3

y

= 3

1

x +

27

3y

+ 3 × 2

1

x ×

3

y + 3 ×

x

1 ×

9

2y

= 3

1

x +

27

3y

+ 2x

y +

x

y

3

2

(ii)

3

2

23

xx

=

33

x –

3

2

2

x – 3 ×

23

x.

2

2

x + 3 ×

x

3 ×

2

2

2

x

= 3

27

x– 6

8

x – 3 × 2

9

x× 2

2

x + 3 ×

x

3× 4

4

x

= 3

27

x – 6

8

x – 4

54

x + 5

36

x

(iii)

33

2

x

x = (2x)3 +

33

x + 3 × (2x)2 ×

x

3 + 3 × 2x ×

23

x

= 8x3 + 3

27

x + 3 × 4x2 ×

x

3 + 3 × 2x × 2

9

x

= 8x3 + 3

27

x + 36x +

x

54

(iv)

3

3

14

x = (4)3 –

3

3

1

x – 3 × (4)2 ×

x3

1



+ 3 × 4 ×

2

3

1

x

= 64 – 327

1

x – 3 × 16 ×

x3

1 + 3 × 4 × 29

1

x

= 64 – 327

1

x –

x

16 + 23

4

x

2. If a + b = 10 and ab = 21, find the value of

a3 + b3.

Sol. a + b = 10, ab = 21

Cubing both sides, (a + b)3 = (10)3

a3 + b3 + 3ab (a + b) = 1000

a3 + b3 + 3 × 21 × 10 = 1000

a3 + b3 + 630 = 1000

a3 + b3 = 1000 – 630 = 370

a3 + b3 = 370

3. If a – b = 4 and ab = 21, find the value of

a3 – b3.

Sol. a – b = 4, ab = 21

Cubing both sides,

(a – b)3 = (4)3

a3 – b3 – 3ab (a – b) = 64

a3 – b3 – 3 × 21 × 4 = 64

a3 – b3 – 252 = 64

a3 – b3 = 64 + 252 = 316

a3 – b3 = 316

4. If x + x


1

x.

Sol. x + x

1 = 5

Cubing both sides,

31

xx = (5)3

x3 + 3

1

x + 3

xx

1 = 125

x3 + 3

1

x + 3 × 5 = 125

x3 + 3

1

x + 15 = 125

x3 + 3

1

x = 125 – 15 = 110

5. If x – x

1 = 7, find the value of x3 – 3

1

x.

Sol. x – x

1 = 7

Cubing both sides,

31

xx = (7)3

x3 – 3

1

x – 3

xx

1 = 343

x3 – 3

1

x – 3 × 7 = 343

x3 – 3

1

x – 21 = 343

x3 – 3

1

x = 343 + 21 = 364

6. If x – x

1 = 5, find the value of x3 – 3

1

x.

Sol. x – x

1 = 5

Cubing both sides,

31

xx = (5)3

x3 – 3

1

x – 3

xx

1 = 125

x3 – 3

1

x – 3 × 5 = 125

x3 – 3

1

x – 15 = 125



x3 – 3

1

x = 125 + 15 = 140

x3 – 3

1

x = 140

7. If x2 + 2

1

x= 51, find the value of x3 – 3

1

x.

Sol. x2 + 2

1

x = 51

21

xx = x2 + 2

1

x – 2

= 51 – 2 = 49 = (7)2

x – x

1 = 7

Cubing both sides,

31

xx = (7)3

x3 – 3

1

x – 3

xx

1 = 343

x3 – 3

1

x – 3 × 7 = 343

x3 – 3

1

x – 21 = 343

x3 – 3

1

x = 343 + 21 = 364

x3 – 3

1

x = 364

8. If x2+ 2

1

x = 98, find the value of x3 + 3

1

x.

Sol.

21

xx = x2 + 2

1

x + 2

= 98 + 2 = 100 = (10)2

x + x

1 = 10

Cubing both sides,

31

xx = (10)3

x3 + 3

1

x + 3

xx

1 = 1000

x3 + 3

1

x + 3 × 10 = 1000

x3 + 3

1

x + 30 = 1000

x3 + 3

1

x = 1000 – 30 = 970

x3 + 3

1

x = 970

9. If 2x + 3y = 13 and xy = 6, find the value

of 8x3 + 27y3.

Sol. 2x + 3y = 13, xy = 6

Cubing both sides,

(2x + 3y)3 = (13)3

(2x)3 + (3y)3 + 3 × 2x × 3y(2x + 3y) = 2197

8x3 + 27y3 + 18xy(2x + 3y) = 2197

8x3 + 27y3 + 18 × 6 × 13 = 2197

8x3 + 27y3 + 1404 = 2197

8x3 + 27y3 = 2197 – 1404 = 793

8x3 + 27y3 = 793

10. If 3x – 2y = 11 and xy = 12, find the value

of 27x3 – 8y3.

Sol. 3x – 2y = 11 and xy = 12

Cubing both sides,

(3x – 2y)3 = (11)3

(3x)3 – (2y)3 – 3 × 3x × 2y(3x – 2y) = 1331

27x3 – 8y3 – 18xy(3x – 2y) = 1331

27x3 – 8y3 – 18 × 12 × 11 = 1331

27x3 – 8y3 – 2376 = 1331

27x3 – 8y3 = 1331 + 2376 = 3707

2x3 – 8y3 = 3707

11. Evaluate each of the following:

(i) (103)3 (ii) (98)3

(iii) (9.9)3 (iv) (10.4)3



(v) (598)3 (vi) (99)3

Sol. We know that (a + b)3 = a3 + b3 + 3ab(a +

b) and (a – b)3 = a3 – b3 – 3ab(a – b)

Therefore,

(i) (103)3 = (100 + 3)3

= (100)3 + (3)3 + 3 × 100 × 3(100 + 3)

{_ (a + b)3 = a3 + b3 + 3ab(a + b)}

= 1000000 + 27 + 900 × 103

= 1000000 + 27 + 92700 = 1092727

(ii) (98)3 = (100 – 2)3

= (100)3 – (2)3 – 3 × 100 × 2(100 – 2)

= 1000000 – 8 – 600 × 98

= 1000000 – 8 – 58800

= 1000000 – 58808 = 941192

(iii) (9.9)3 = (10 – 0.1)3

= (10)3 – (0.1)3 – 3 × 10 × 0.1(10 – 0.1)

= 1000 – 0.001 – 3 × 9.9

= 1000 – 0.001 – 29.7

= 1000 – 29.701 = 970.299

(iv) (10.4)3 = (10 + 0.4)3

= (10)3 + (0.4)3 + 3 × 10 × 0.4(10 + 0.4)

= 1000 + 0.064 + 12(10.4)

= 1000 + 0.064 + 124.8 = 1124.864

(v) (598)3 = (600 – 2)3

= (600)3 – (2)3 – 3 × 600 × 2 × (600 – 2)

= 216000000 – 8 – 3600 × 598

= 216000000 – 8 – 2152800

= 216000000 – 2152808 = 213847192

(vi) (99)3 = (100 – 1)3

= (100)3 – (1)3 – 3 × 100 × 1 × (100 – 1)

= 1000000 – 1 – 300 × 99

= 1000000 – 1 – 29700

= 1000000 – 29701 = 970299

12. Evaluate each of the following:

(i) 1113 – 893 (ii) 463 + 343

(iii) 1043 + 963 (iv) 933 – 1073

Sol. We know that a3 + b3 = (a + b)3 – 3ab(a +

b) and a3 – b3 = (a – b)3 + 3ab(a – b)

(i) 1113 – 893

= (111 – 89)3 + 3 × 111 × 89(111 – 89)

= (22)3 + 3 × 111 × 89 × 22

= 10648 + 652014 = 662662

OR

(a + b)3 – (a – b)3 = 2(b3 + 3a2b)

= 1113 – 893 = (100 + 11)3 – (100 – 11)3

= 2[113 + 3 × 1002 × 11]

= 2[1331 + 330000]

= 331331 × 2 = 662662

(ii) (a + b)3 + (a – b)3 = 2(b3 + 3ab2)

463 + 343 = (40 + 6)3 + (40 – 6)3

= 2[(40)3 + 3 × 40 × 62]

= 2[64000 + 3 × 40 × 36] = 2[64000 +

4320]

= 2 × 68320 = 136640

(iii) 1043 + 963 = (100 + 4)3 + (100 – 96)3

= 2[a3 + 3ab2]

= 2[(100)3 + 3 × 100 × (4)2]

= 2[1000000 + 300 × 16]

= 2[1000000 + 4800]

= 1004800 × 2 = 2009600

(iv) 933 – 1073 = –[(107)3 – (93)3]

= –[(100 + 7)3 – (100 – 7)3]

= –2[b3 + 3a2b] = –2[(7)3 + 3(100)2 × 7]

= –2[343 + 3 × 10000 × 7]

= –2[343 + 210000]

= –2[210343] = –420686

13. If x + x

1 = 3, calculate x2 + 2

1

x, x3 + 3

1

x

and x4 + 4

1

x.

Sol. (i) x + x

1 = 3


21

xx = (3)2 x2 + 2

1

x + 2 = 9

x2 + 2

1

x = 9 – 2 = 7

x2 + 2

1

x = 7



(ii) x + x

1 = 3

Cubing both sides,

31

xx = (3)3

x3 + 3

1

x + 3

xx

1 = 27

x3 + 3

1

x + 3 × 3 = 27 x3 + 3

1

x + 9 = 27

x3 + 3

1

x = 27 – 9 = 18

x3 + 3

1

x = 18

(iii) x2 + 2

1

x = 7


2

2 1

x

x = (7)2

x4 + 4

1

x + 2 = 49

x4 + 4

1

x = 49 – 2 = 47

x4 + 4

1

x = 47

14. Find the value of 27x3 + 8y3, if

(i) 3x + 2y = 14 and xy = 8

(ii) 3x + 2y = 20 and xy = 9

14

Sol. (i) 3x + 2y = 14 and xy = 8

Cubing both sides,

(3x + 2y)3 = (14)3

(3x)3 + (2y)3 + 3 × 3x × 2y(3x + 2y) = 2744

27x3 + 8y3 + 18xy(3x + 2y) = 2744

27x3 + 8y3 + 18 × 8 × 14 = 2744

27x3 + 8y3 + 2016 = 2744

27x3 + 8y3 = 2744 – 2016 = 728

(ii) 3x + 2y = 20 and xy = 9

14

Cubing both sides,

(3x + 2y)3 = (20)3

(3x)3 + (2y)3 + 3 × 3x × 2y(3x + 2y) = 8000

27x3 + 8y3 + 3 × 3 × 2xy(3x + 2y) = 8000

27x3 + 8y3 + 18 × 9

14 × 20 = 8000

27x3 + 8y3 + 560 = 8000

27x3 + 8y3 = 8000 – 560 = 7440

27x3 + 8y3 = 7440

15. Find the value of 64x3 – 125z3, if 4x – 5z =

16 and xz = 12.

Sol. 4x – 5z = 16, xz = 12

Cubing both sides,

(4x – 5z)3 = (16)3

(4x)3 – (5y)3 – 3 × 4x × 5z(4x – 5z) = 4096

64x3 – 125z3 – 3 × 4 × 5 × xz(4x – 5z) = 4096

64x3 – 125z3 – 60 × 12 × 16 = 4096

64x3 – 125z3 – 11520 = 4096

64x3 – 125z3 = 4096 + 11520 = 15616

16. If x – x

1 = 3 + 2 2 , find the value of x3 –

3

1

x.

Sol. x – x

1 = 3 + 2 2

Cubing both sides,

31

xx = (3 + 2 2 )3

x3 – 3

1

x – 3

xx

1 = (3)3 + (2 2 )3 + 3

× 3 × 2 2 (3 + 2 2 )



x3 – 3

1

x – 3 × (3 + 2 2 ) = 27 + 16 2 +

18 2 (3 + 2 2 )

x3 – 3

1

x – 3 × (3 + 2 2 ) = 27 + 16 2 +

54 2 + 72

x3 + 3

1

x – 3(3 + 2 2 ) = 99 + 70 2

x3 + 3

1

x = 99 + 70 2 + 3(3 + 2 2 )

= 99 + 70 2 + 9 + 6 2

= 108 + 76 2


(i) (x + 3)3 + (x – 3)3

(ii)

3

32

yx

–

3

32

yx

(iii)

32

xx +

32

xx

(iv) (2x – 5y)3 – (2x + 5y)3

Sol. (i) (x + 3)3 + (x – 3)3

= (x)3 + (3)3 + 3x2 × 3 + 3 × x × (3)2 +

[(x)3 – (3)3 – 3x2 × 3 + 3 × x (3)2]

= x3 + 27 + 9x2 + 27x + x3 – 27 – 9x2 + 27x

= 2x3 + 54x

(ii)

3

32

yx

–

3

32

yx

=

3

2

x +

3

3

y + 3 ×

2

2

x

× 3

y +

3

2

x

2

3

y –

3

2

x –

3

3

y – 3

2

2

x

× 3

y + 3

2

x

2

3

y

=

8

3x +

27

3y

+ 3 × 4

2x

× 3

y + 3 ×

2

x ×

9

2y –

8

3x –

27

3y

– 3 × 4

2x

× 3

y + 3

2

x

×

9

2y

= 8

3x

+ 27

3y

+ 4

2yx

+ 6

2xy

– 8

3x

+ 27

3y

+ 4

2yx

– 6

2xy

= 27

23

y +

2

2yx

(iii)

32

xx +

32

xx

=

3

3 8

xx + 3 × x2 ×

x

2 + 3x ×

2

4

x +

3

3 8

xx – 3x2 ×

x

2 + 3x ×

2

4

x

= x3 + 3

8

x + 6x +

x

12 + x3 – 3

8

x – 6x +

x

12

= 2x3 + x

24

(iv) (2x – 5y)3 – (2x + 5y)3

= [(2x)3 – (5y)3 – 3(2x)2 (5y) + 3 × 2x ×

(5y)2] – [(2x)3 + (5y)3 + 3(2x)2 (5y) + 3 ×

2x × (5y)2]

= [8x3 – 125y3 – 60x2y + 150xy2] – [8x3 +

125y3 + 60x2y + 150xy2]

= 8x3 – 125y3 – 60x2y + 150xy2 – 8x3 –

125y3 – 60x2y – 150xy2

= –250y2 – 120x2y

18. If x4 + 4

1

x = 194, find x3 + 3

1

x, x2 + 2

1

x



and x + x

1.

Sol. x4 + 4

1

x = 194

Adding 2 to both sides,

x4 + 4

1

x + 2 = 194 + 2 = 196

(x2)2 + 22 )(

1

x + 2 = (14)2

2

2

2 1

x

x = (14)2 x2 + 2

1

x = 14

x2 + 2

1

x = 14


x2 + 2

1

x + 2 = 14 + 2 = 16 = (4)2

21

xx = (4)2

x + x

1 = 4

Cubing both sides,

31

xx = x3 + 3

1

x + 3

xx

1

(4)3 = x3 + 3

1

x + 3 × 4

64 = x3 + 3

1

x + 12

x3 + 3

1

x = 64 – 12 = 52

x3 + 3

1

x = 52, x2 + 2

1

x = 14

and x + x

1 = 4

19. If x4 + 4

1

x= 119, find the value of x3 – 3

1

x.

Sol. x4 + 4

1

x = 119


x4 + 4

1

x + 2 = 119 + 2

(x2)2 +

2

2

1

x+ 2 = 121

2

2

2 1

x

x = (11)2 x2 + 2

1

x = 111

x2 + 2

1

x = 111

Subtracting 2 from both sides,

x2 + 2

1

x – 2 = 11 – 2 = 9

2

1

xx = (3)2 x –

x

1 = 3

x – x

1 = 3

Cubing both sides,

31

xx = (3)3

x3 – 3

1

x – 3

xx

1 = 27

x3 – 3

1

x – 3 × 3 = 27 x3 – 3

1

x – 9 = 27

x3 – 3

1

x = 27 + 9 = 36

EXERCISE 4.4

1. Find the following products:

(i) (3x + 2y) (9x2 – 6xy + 4y2)



(ii) (4x – 5y) (16x2 + 20xy + 25y2)

(iii) (7p4 + q) (49p8 – 7p4q + q2)

(iv)

yx

22

2

2

44

yxyx

(v)

yx

53

xyyx

1525922

(vi)

x

53

2

25159

xx

(vii)

xx

32

694 2

2x

x

(viii)

22

3x

x

xxx

649 4

2

(ix) (1 – x) (1 + x + x2)

(x) (1 + x) (1 – x + x2)

(xi) (x2 – 1) (x4 + x2 + 1)

(xii) (x3 + 1) (x6 – x3 + 1)

Sol. We know that a3 + b3 = (a + b) (a2 – ab +

b2) and a3 – b3 = (a – b) (a2 + ab + b2)

(i) (3x + 2y) (9x2 – 6xy + 4y2)

= (3x + 2y) [(3x)2 – 3x × 2y + (2y)2]

= (3x)3 + (2y)3 = 27x3 + 8y3

(ii) (4x – 5y) (16x2 + 20xy + 25y2)

= (4x – 5y) [(4x)2 + 4x × 5y + (5y)2]

= (4x)3 – (5y)3

= 64x3 – 125y3

(iii) (7p4 + q) (49p8 – 7p4q + q2)

= (7p4 + q) [(7p4)2 – 7p4 × q + (q)2]

= (7p4)3 + (q)3 = 343p12 + q3

(iv)

yx

22

2

2

44

yxyx

=

yx

22

2

2

)2(222

yyxx

=

3

2

x

+ (2y)3 = 8

3x

+ 8y3

(v)

yx

53

xyyx

1525922

=

yx

53

225533

yyxx

=

33

x–

35

y

= 3

27

x – 3

125

y

(vi)

x

53

2

25159

xx

=

x

53

2

2 553)3(

xx

= (3)3 +

35

x= 27 + 2

125

x

(vii)

xx

32

694 2

2x

x

=

xx

32

2

296

4x

x

=

xx

32

2

2

)3(322

xxxx

=

32

x+ (3x)3 = 3

8

x + 27x3

(viii)

222

3x

xxx

649 4

2

=

222

3x

4

246

9xx

x

=

222

3x

222

2

)2(223

xxxx



=

33

x– (2x2)3 = 3

27

x – 8x6

(ix) (1 – x) (1 + x + x2)

= (1 – x) [(1)2 + 1 × x + (x)2]

= (1)3 – (x)3 = 1 – x3

(x) (1 + x) (1 – x + x2)

= (1 + x) [(1)2 – 1 × x + (x)2]

= (1)3 + (x)3 = 1 + x3

(xi) (x2 – 1) (x4 + x2 + 1)

= (x2 – 1) [(x2)2 + x2 × 1 + (1)2]

= (x2)3 – (1)3

= x6 – 1

(xii) (x3 + 1) (x6 – x3 + 1)

= (x3 + 1) [(x3)2 – x3 × 1 + (1)2]

= (x3)3 + (1)3 = x9 + 1

2. If x = 3 and y = –1, find the values of each

of the following using in identity:

(i) (9y2 – 4x2) (81y4 + 36x2y2 + 16x4)

(ii)

3

3 x

x

1

9

9 2

2

x

x

(iii)

37

yx

21949

22 xyyx

(iv)

34

yx

91216

22 yxyx

(v)

xx

55

2

22525

25x

x

Sol. x = 3, y = –1

(i) (9y2 – 4x2) (81y4 + 36x2y2 + 16x4)

= (9y2 – 4x2) [(9y2)2 + 9y2 × 4x2 + (4x2)2]

= (9y2)3 – (4x2)3 = 729y6 – 64x6

= 729 × (–1)6 – 64(3)6

= 729 × 1 – 64 × 729

= 729 – 46656 = –45927

(ii)

3

3 x

x

1

9

9 2

2

x

x

=

3

3 x

x

22

33

33 xx

xx

=

33

x–

3

3

x

=

3

3

3

–

3

3

3

= 13 – 13 = 0

(iii)

37

yx

21949

22 xyyx

=

37

yx

22

3777

yyxx

=

3

7

x

+

3

3

y

= 343

3x

+ 27

3y

= 343

)3( 3

+ 27

)1( 3

= 343

27 –

27

1

= 9261

343729 =

9261

386

(iv)

34

yx

91216

22 yxyx

=

34

yx

22

3344

yyxx

=

3

4

x

–

3

3

y

= 64

3x

– 27

3y

= 64

)3(3

– 27

)1(3

= 64

27 +

27

1



= 1728

64729 =

1728

793

(v)

xx

55

2

22525

25x

x

=

xx

55

2

2

)5(555

xxxx

=

35

x+ (5x)3 = 3

125

x + 125x3

= 3)3(

125 + 125 × (3)3 =

9

125 + 125 × 27

= 27

125 + 3375

= 27

91125125 =

27

91250


a2 – ab + b2 and a2 + ab + b2.

Sol. a + b = 10, ab = 16

Squaring,

(a + b)2 = (10)2

a2 + b2 + 2ab = 100

a2 + b2 + 2 × 16 = 100

a2 + b2 + 32 = 100

a2 + b2 = 100 – 32 = 68

Now, a2 – ab + b2 = a2 + b2 – ab

= 68 – 16 = 52

and a2 + ab + b2 = a2 + b2 + ab

= 68 + 16 = 84


a3 + b3.

Sol. a + b = 8, ab = 6

Cubing both sides,

(a + b)3 = (8)3

a3 + b3 + 3ab(a + b) = 512

a3 + b3 + 3 × 6 × 8 = 512

a3 + b3 + 144 = 512

a3 + b3 = 512 – 144 = 368

a3 + b3 = 368


a3 – b3.

Sol. a – b = 6, ab = 20

Cubing both sides,

(a – b)3 = (6)3

a3 – b3 – 3ab(a – b) = 216

a3 – b3 – 3 × 20 × 6 = 216

a3 – b3 – 360 = 216

a3 – b3 = 216 + 360 = 576

a3 – b3 = 576

6. If x = –2 and y = 1, by using an identity

find the value of the following:

(i) (4y2 – 9x2) (16y4 + 36x2y2 + 81x4)

(ii)

2

2 x

x

1

4

4 2

2

x

x

(iii)

yy

155

2

2 2257525

yy

Sol. x = –2, y = 1

(i) (4y2 – 9x2) (16y4 + 36x2y2 + 81x4)

= (4y2 – 9x2) [(4y2)2 + 4y2 × 9x2 + (9x2)2]

= (4y2)3 – (9x2)3 = 64y6 – 729x6

= 64 × 16 – 729 × (–2)6

= 64 × 1 – 729 × (64)

= 64 – 46656 = –46592

(ii)

2

2 x

x

1

4

4 2

2

x

x

=

2

2 x

x

22

22

22

2 xx

xx

=

32

x–

3

2

x

= 3

8

x –

8

3x

= 3)2(

8

– 8

)2(3

= 8

8

–

8

8

= –1 + 1 = 0



(iii)

yy

155

2

2 2257525

yy

=

yy

155

2

2 151552)5(

yyyy

= (5y)3 +

315

y= 125y3 + 3

3375

y

= 125 × (1) + 3)1(

3375

= 125 + 1

3375 = 125 + 3375 = 3500

EXERCISE 4.5

1. Find the following products:

(i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy – 4yz

– 6zx)

(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy +

6yz – 8zx)

(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab –

6bc + 4ca)

(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy –

15zx + 20yz)

Sol. (i) (3x + 2y + 2z) (9x2 + 4y2 + 4z2 – 6xy –

4yz – 6zx)

= (3x + 2y + 2z) [(3x)2 + (2y)2 + (2z)2 – 3x

× 2y + 2y × 2z + 2z × 3x]

= (3x)3 + (2y)3 + (2z)3 – 3 × 3x × 2y × 2z

= 27x3 + 8y3 + 8z3 – 36xyz

(ii) (4x – 3y + 2z) (16x2 + 9y2 + 4z2 + 12xy +

6yz – 8zx)

= (4x – 3y + 2z) [(4x)2 + (–3y)2 + (2z)2 –

4x × (–3y) + (3y) × (2z) – (2z × 4x)]

= (4x)3 + (–3y)3 + (2z)3 – 3 × 4x × (–3y) × (2z)

= 64x3 – 27y3 + 8z3 + 72xyz

(iii) (2a – 3b – 2c) (4a2 + 9b2 + 4c2 + 6ab –

6bc + 4ca)

= (2a – 3b – 2c) [(2a)2 + (3b)2 + (2c)2 – 2a

× (–3b) – (–3b) × (–2c) – (–2c) × 2a]

= (2a)3 + (3b)3 + (–2c)3 – 3 × 2a × (–3b)

(–2c)

= 8a3 – 27b3 – 8c3 – 36abc

(iv) (3x – 4y + 5z) (9x2 + 16y2 + 25z2 + 12xy –

15zx + 20yz)

= [3x + (–4y) + 5z] [(3x)2 + (–4y)2 + (5z)2

– 3x × (–4y) – (–4y) (5z) – 5z × 3x]

= (3x)3 + (–4y)3 + (5z)3 – 3 × 3x × (–4y) (5z)

= 27x3 – 64y3 + 125z3 + 180xyz

2. Evaluate:

(i) 253 – 753 + 503 (ii) 483 – 303 – 183

(iii)

3

2

1

+

3

3

1

–

3

6

5

(iv) (0.2)3 – (0.3)3 + (0.1)3

Sol. (i) We know that if a + b + c = 0, then

x3 + y3 + z3 = 3xyz. Therefore,

(i) (25)3 – (75)3 + (50)3

Let 25 = a, –75 = b and 50 = c

a + b + c = 25 – 75 + 50 = 0

a3 + b3 + c3 = 3abc

253 – 753 + 503 = 3 × 25 × (–75) × 50

= –3 × 25 × 75 × 50 = –281250

(ii) 483 – 303 – 183

Let a = 48, b = –30, c = –18

a + b + c = 48 – 30 – 18 = 0, then

a3 – b3 – c3 = 3abc

= 3 × 48 × (–30) × (–18)

= 3 × 48 × 30 × 18 = 77760

(iii)

3

2

1

+

3

3

1

–

3

6

5

Let a = 2

1, b =

3

1, c =

6

5

a + b + c = 2

1 +

3

1 –

6

5

= 6

523 =

6

0 = 0

a3 + b3 + c3 = 3abc



3

2

1

+

3

3

1

–

3

6

5

= 3 × 2

1 ×

3

1 ×

6

5

= –3 × 2

1 ×

3

1 ×

6

5 =

12

5

(iv) (0.2)3 – (0.3)3 + (0.1)3

Let a = 0.2, b = –0.3, c = 0.1

a + b + c = 0.2 – 0.3 + 0.1 = 0

a3 + b3 + c3 = 3abc

(0.2)3 – (0.3)3 + (0.1)3

= 3(0.2) (–0.3) (0.1)

= –3 × 0.006 = –0.018

3. If x + y + z = 8 and xy + yz + zx = 20, find

the value of x3 + y3 + z3 – 3xyz.

Sol. We know that

x3 + y3 + z3 – 3xyz = (x + y + z) (x2 + y2 +

z2 – xy – yz – zx)

Now, x + y + z = 8

Squaring, we get

(x + y + z)2 = (8)2

x2 + y2 + z2 + 2(xy + yz + zx) = 64

x2 + y2 + z2 + 2 × 20 = 64

x2 + y2 + z2 + 40 = 64

x2 + y2 + z2 = 64 – 40 = 24

Now,

x3 + y3 + z3 – 3xyz = (x + y + z) [x2 + y2 +

z2 – (xy + yz + zx)]

= 8(24 – 20) = 8 × 4 = 32

4. If a + b + c = 9 and ab + bc + ca = 26,

find the value of a3 + b3 + c3 – 3abc.

Sol. a + b + c = 9, ab + bc + ca = 26

Squaring, we get

(a + b + c)2 = (9)2

a2 + b2 + c2 + 2(ab + bc + ca) = 81

a2 + b2 + c2 + 2 × 26 = 81

a2 + b2 + c2 + 52 = 81

a2 + b2 + c2 = 81 – 52 = 29

Now, a3 + b3 + c3 – 3abc

= (a + b + c) [(a2 + b2 + c2 – (ab + bc + ca)]

= 9[29 – 26] = 9 × 3 = 27

5. If a + b + c = 9, and a2 + b2 + c2 = 35, find

the value of a3 + b3 + c3 – 3abc.

Sol. a + b + c = 9

Squaring, we get

(a + b + c)2 = (9)2

a2 + b2 + c2 + 2(ab + bc + ca) = 81

35 + 2(ab + bc + ca) = 81

2(ab + bc + ca) = 81 – 35 = 46

ab + bc + ca = 2

46 = 23

Now, a3 + b3 + c3 – 3abc

= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]

= 9[35 – 23] = 9 × 12 = 108


(VSAQs)

1. If x + x

1 = 3, then find the value of x2 +

2

1

x.

Sol. x + x

1 = 3


21

xx = (3)2

x2 + 2

1

x + 2 = 9 x2 + 2

1

x = 9 – 2 = 7

x2 + 2

1

x = 7

2. If x + x


1

x.

Sol. x + x

1 = 3,


21

xx = (3)2



x2 + 2

1

x + 2 = 9

x2 + 2

1

x = 9 – 2 = 7

Cubing both sides,

3

2

2 1

x

x = (7)3

x6 + 6

1

x + 3

2

2 1

xx = 343

x6 + 6

1

x + 3 × 7 = 343

x6 + 6

1

x + 21 = 343

x6 + 6

1

x = 343 – 21 = 322

x6 + 6

1

x = 322


a2 + b2.

Sol. a + b = 7, ab = 12


(a + b)2 = (7)2

a2 + b2 + 2ab = 49

a2 + b2 + 2 × 12 = 49

a2 + b2 + 24 = 49

a2 + b2 = 49 – 24 = 25

a2 + b2 = 25


a2 + b2.

Sol. a – b = 5, ab = 12


(a – b)2 = (5)2

a2 + b2 – 2ab = 25

a2 + b2 – 2 × 12 = 25

a2 + b2 – 24 = 25

a2 + b2 = 25 + 24 = 49

a2 + b2 = 49

5. If x – x

1 =

2

1, then write the value of 4x2

+ 2

4

x.

Sol. x – x

1 =

2

1

2x – x

2 = 1 (Multiplying by 2)


22

2

x

x = (1)2

4x2 + 2

4

x – 2 × 2x ×

x

2 = 1

4x2 + 2

4

x – 8 = 1

4x2 + 2

4

x = 1 + 8 = 9

4x2 + 2

4

x = 9

6. If a2 + 2

1

a = 102, find the value of a –

a

1.

Sol. a2 + 2

1

a = 102

21

aa = a2 + 2

1

a – 2

= 102 – 2 = 100 = (10)2

a – a

1 = 10

7. If a + b + c = 0, then write the value of

bc

a2

+ ca

b2

+ ab

c2

.

Sol.bc

a2

+ ca

b2

+ ab

c2



= abc

cba333

a + b + c = 0

Then, a3 + b3 + c3 = 3abc

abc

cba333

= abc

abc3 = 3



following:

1. If x + x

1 = 5, then x2 + 2

1

x =

(a) 25 (b) 10

(c) 23 (d) 27

Sol. x + x

1 = 5

Squaring, we get

x2 + 2

1

x + 2 = 25

x2 + 2

1

x = 25 – 2 = 23 (c)

2. If x + x

1 = 2, then x3 + 3

1

x =

(a) 64 (b) 14

(c) 8 (d) 2

Sol. x + x

1 = 2,

Cubing, we get

x3 + 3

1

x + 3

2

1x = (2)3

x3 + 3

1

x × 3(2) = 8

x3 + 3

1

x = 8 – 6 = 2 (d)

3. If x + x

1 = 4, then x4 + 4

1

x =

(a) 196 (b) 194

(c) 192 (d) 190

Sol. x + x

1 = 4

Squaring, we get

x2 + 2

1

x + 2 = 16

x2 + 2

1

x = 16 – 2 = 14

Again squaring,

x4 + 4

1

x + 2 = 196

x4 + 4

1

x = 196 – 2 = 194 (b)

4. x + x

1 = 3, then x6 + 6

1

x =

(a) 927 (b) 414

(c) 364 (d) 322

Sol. x + x

1 = 3

Squaring, we get

x2 + 2

1

x + 2 = 9

x2 + 2

1

x = 9 – 2 = 7

Cubing both sides,

3

2

2 1

x

x = (7)3

x6 + 6

1

x + 3

2

2 1

xx = 343

x6 + 6

1

x + 3 × 7 = 343



x6 + 6

1

x + 21 = 343

x6 + 6

1

x = 343 – 21 = 322 (d)

5. If x2 + 2

1

x = 102, then x –

x

1 =

(a) 8 (b) 10

(c) 12 (d) 13

Sol. x2 + 2

1

x = 102

Subtracting 2 from both sides

x2 + 2

1

x – 2 = 102 – 2 = 100

21

xx = (10)2

x – x

1 = 10 (b)

6. If x3 + 3

1

x = 110, then x +

x

1 =

(a) 5 (b) 10


Sol. x3 + 3

1

x = 110

x3 + 3

1

x =

31

xx – 3

xx

1

110 =

31

xx – 3

xx

1

Let x + x

1 = a, then

a3 – 3a – 110 = 0

Factors of –110 = +1, +2, +5, +11

Let a = 1, then 1 – 3 – 110 0

Let a = –1, then –1 + 3 – 110 0

Let a = 2, then 8 – 6 – 110 0

Let a = –2, then –8 – 6 – 110 0

Let a = 5, then 125 – 15 – 110 = 0

a = 5 x + x

1 = 5 (a)

7. If x3 – 3

1

x = 14, then x +

x

1 =

(a) 5 (b) 4

(c) 3 (d) 2

Sol.

31

xx = x3 –

x

1 – 3

xx

1

31

xx = 14 – 3

xx

1

Let x – x

1 = a, then

a3 = 14 – 3a a3 + 3a – 14 = 0

Factors of –14 = +1, +2, +7

Let a= +1, then 1 + 3 – 14 0

Let a = –1, then –1 – 3 – 14 0

Let a = 2, then 8 + 6 – 14 = 0

x – x

1 = 2 (d)

8. If a + b + c = 9 and ab + bc + ca = 23,

then a2 + b2 + c2 =

(a) 35 (b) 58


Sol. a + b + c = 9, ab + bc + ca = 23

Squaring,

(a + b + c)2 = (9)2

a2 + b2 + c2 + 2(ab + bc + ca) = 81

a2 + b2 + c2 + 2 × 23 = 81

a2 + b2 + c2 + 46 = 81

a2 + b2 + c2 = 81 – 46 = 35 (a)

9. (a – b)3 + (b – c)3 + (c – a)3 =

(a) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

(b) (a – b) (b – c) (c – a)

(c) 3(a – b) (b – c) (c – a)

(d) none of these



Sol. (a – b)3 + (b – c)3 + (c – a)3

_ a – b + b – c + c – a = 0

(a – b)3 + (b – c)3 + (c – a)3

= 3(a – b) (b – c) (c – a) (c)

10. If b

a +

a

b = 1, then a3 + b3 =

(a) 1 (b) –1

(c)2

1(d) 0

Sol.b

a +

a

b = 1

ab

ba22

= 1 a2 + b2 = ab

Now, a3 + b3 = (a + b) (a2 – ab + b2)

= (a + b) (a2 + b2 – ab)

= (a + b) (ab – ab)

= (a + b) × 0 = 0 (d)

11. If a – b = –8 and ab = –12 then a3 – b3 =

(a) –244 (b) –240

(c) –224 (d) –260

Sol. a – b = –8, ab = –12

(a – b)3 = a3 – b3 – 3ab (a – b)

(–8)3 = a3 – b3 – 3 × (–12) (–8)

–512 = a3 – b3 – 288

a3 – b3 = –512 + 288 = –224 (c)

12. If the volume of a cuboid is 3x2 – 27, then

its possible dimensions are

(a) 3, x2, –27x (b) 3, x – 3, x + 3

(c) 3, x2, 27x (d) 3, 3, 3

Sol. Volume = 3x2 – 27 = 3(x2 – 9)

= 3(x + 3) (x – 3)

Dimensions are = 3, x – 3, x + 3 (b)

13. 75 × 75 + 2 × 75 × 25 + 25 × 25 is equal

to

(a) 10000 (b) 6250

(c) 7500 (d) 3750

Sol. 75 × 75 + 2 × 75 × 25 + 25 × 25

= (75)2 + 2 × 75 × 25 + (25)2

{_ a2 + 2ab + b2 = (a + b)2}

= (75 + 25)2 = (100)2

= 10000 (a)

14. (x – y) (x + y) (x2 + y2) (x4 + y4) is equal to

(a) x16 – y16 (b) x8 – y8

(c) x8 + y8 (d) x16 + y16

Sol. (x – y) (x + y) (x2 + y2) (x4 + y4)

= (x2 – y2) (x2 + y2) (x4 + y4)

{_ (a + b) (a – b) = a2 – b2}

= [(x2)2 – (y2)2] (x4 + y4) = (x4 – y4) (x4 + y4)

= (x4)2 – (y4)2 = x8 – y8 (b)

15. If x4 + 4

1

x = 623, then x +

x

1 =

(a) 27 (b) 25

(c) 3 3 (d) –3 3

Sol. x4 + 4

1

x = 623


x4 + 4

1

x + 2 = 623 + 2 = 625

2

2

2 1

x

x = (25)2

x2 + 2

1

x = 25


x2 + 2

1

x + 2 = 25 + 2 = 27

21

xx = ( 27 )2 = 239

x + x

1 = 3 3 (c)

16. If x4 + 4

1

x = 194, then x3 + 3

1

x =

(a) 76 (b) 52




Sol. x4 + 4

1

x = 194


x4 + 4

1

x + 2 = 194 + 2 = 196

2

2

2 1

x

x = (14)2 x2 + 2

1

x = 14


x2 + 2

1

x + 2 = 14 + 2 = 16

2

1

xx = (4)2

x + x

1 = 4

Cubing both sides,

31

xx = (4)3

x3 + 3

1

x + 3

xx

1 = 64

x3 + 3

1

x + 3 × 4 = 64

x3 + 3

1

x + 12 = 64

x3 + 3

1

x = 64 – 12 = 52 (b)

17. If x – x

1 =

4

15, then x +

x

1 =

(a) 4 (b)4

17

(c)4

13(d)

4

1

Sol. x – x

1 =

4

15

21

xx =

21

xx + 4

=

2

4

15

+ 4 = 16

225 + 4 =

16

64225

= 16

289 =

2

4

17

x + x

1 =

4

17(b)

18. If 3x + x

2 = 7, then

2

2 49

xx =

(a) 25 (b) 35

(c) 49 (d) 30

Sol. 3x + x

2 = 7

Squaring, we get

22

3

x

x = (7)2 = 49

9x2 + 2

4

x + 2 × 3x ×

x

2 = 49

9x2 + 2

4

x + 12 = 49

9x2 + 2

4

x = 49 – 12 = 37

22

3

x

x = 9x2 + 2

4

x – 2 × 3x ×

x

2

= 9x2 + 2

4

x – 12 = 37 – 12 = 25 = (5)2

3x – x

2 = 5



Now,

x

x2

3

x

x2

3 = 7 × 5

(3x)2 –

22

x= 35

9x2 – 2

4

x = 35 (b)

19. If a2 + b2 + c2 – ab – bc – ca = 0, then

(a) a + b = c (b) b + c = a

(c) c + a = b (d) a = b = c

Sol. a2 + b2 + c2 – ab – bc – ca = 0

2(a2 + b2 + c2 – ab – bc – ca) = 0

(Multiplying by 2)

2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 0

a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 –

2ca = 0

(a – b)2 + (b – c)2 + (c – a)2 = 0

(a – b)2 = 0, then a – b = 0 a = b

Similarly, (b – c)2 = 0, then

b – c = 0 b = c

and (c – a)2 = 0, then c – a = 0 c = a

a = b = c (d)

20. If a + b + c = 0, then bc

a2

+ ca

b2

+ ab

c2

=

(a) 0 (b) 1

(c) –1 (d) 3

Sol. a + b + c = 0

bc

a2

+ ca

b2

+ ab

c2

= abc

a3

+ abc

b3

+ abc

c3

= abc

cba333

_ a + b + c = 0

Then, a3 + b3 + c3 = 3abc

= abc

abc3 = 3 (d)

21. If 3

1

a + 3

1

b + 3

1

c = 0, then

(a) a + b + c = 0

(b) (a + b + c)3 = 27abc

(c) a + b + c = 3abc

(d) a3 + b3 + c3 = 0

Sol. _ 3

1

a + 3

1

b + 3

1

c = 0

3

3

1

a +

3

3

1

b +

3

3

1

c = 3 3

1

a 3

1

b 3

1

c

a + b + c = 3 3

1

a 3

1

b 3

1

c

(a + b + c)3 = (3 3

1

a 3

1

b 3

1

c )3

(Cubing both sides)

(a + b + c)3 = 27abc (b)

22. If a + b + c = 9 and ab + bc + ca = 23,

then a3 + b3 + c3 – 3abc =

(a) 108 (b) 207

(c) 669 (d) 729

Sol. a3 + b3 + c3 – 3abc

= (a + b + c) [a2 + b2 + c2 – (ab + bc + ca)]

Now, a + b + c = 9

Squaring,

a2 + b2 + c2 + 2(ab + bc + ca) = 81

a2 + b2 + c2 + 2 × 23 = 81

a2 + b2 + c2 + 46 = 81

a2 + b2 + c2 = 81 – 46 = 35

Now, a3 + b3 + c3 – 3abc = (a + b + c) [(a2

+ b2 + c2) – (ab + bc + ca)]

= 9[35 – 23] = 9 × 12 = 108 (a)

23. 333

322322322

)()()(

)()()(

accbba

accbba

=

(a) 3(a + b) (b + c) (c + a)

(b) 3(a – b) (b – c) (c – a)

(c) (a – b) (b – c) (c – a)

(d) (a + b) (b + c) (c + a)

Sol. a2 – b2 + b2 – c2 + c2 – a2 = 0

(a2 – b2)3 + (b2 – c2)3 + (c2 – a2)3

= 3(a2 – b2) (b2 – c2) (c2 – a2)



and a – b + b – c + c – a = 0

(a – b)3 + (b – c)3 + (c – a)3

= 3(a – b) (b – c) (c – a)

Now, 333

322322322

)()()(

)()()(

accbba

accbba

= )()()(3

)()()(3 222222

accbba

accbba

= )()()(3

)()()()()()(3

accbba

acaccbcbbaba

= (a + b) (b + c) (c + a) (a)

24. The product (a + b) (a – b) (a2 – ab + b2)

(a2 + ab + b2) is equal to

(a) a6 + b6 (b) a6 – b6

(c) a3 – b3 (d) a3 + b3

Sol. (a + b) (a – b) (a2 – ab + b2) (a2 + ab + b2)

= (a + b) (a2 – ab + b2) (a – b) (a2 + ab + b2)

= (a3 + b3) (a3 – b3) = (a3)2 – (b3)2

= a6 – b6 (b)

25. The product (x2 – 1) (x4 + x2 + 1) is equal

to

(a) x8 – 1 (b) x8 + 1

(c) x6 – 1 (d) x6 + 1

Sol. (x2 – 1) (x4 + x2 + 1)

= (x2)3 – (1)3 = x6 – 1 (c)

26. If b

a +

a

b = 1, then a3 + b3 =

(a) 1 (b) –1

(c)2

1(d) 0

Sol.b

a +

a

b = 1

ab

ba22

= 1

a2 + b2 = –ab

a2 + b2 + ab = 0 a2 + ab + b2 = 0

Now, a3 – b3 = (a – b) (a2 + ab + b2)

= (a – b) × 0 = 0 (d)

27. If 49a2 – b =

2

17a

2

17a , then the

value of b is

(a) 0 (b)4

1

(c)2

1(d)

2

1

Sol. 49a2 – b = (7a)2 – ( b )2

= (7a + b ) (7a – b )

(7a + b ) (7a – b ) =

2

17a

2

17a

Comparing,

b = 2

1 b =

4

1(b)




We can factorize.

1. By taking common

2. By using identities

(a + b)2 = a2 + 2ab + b2

(a – b)2 = a2 – 2ab + b2

a2 – b2 = (a + b) (a – b)

a3 + b3 = (a + b) (a2 – ab + b2)

a3 – b3 = (a – b) (a2 + ab + b2)

(x + a) (x + b) = x2 + (a + b)x + ab

EXERCISE 5.1

Factorize

1. x3 + x – 3x2 – 3

Sol. x3 + x – 3x2 – 3

x3 – 3x2 + x – 3

x2(x – 3) + 1(x – 3)

= (x – 3) (x2 + 1)

2. a(a + b)3 – 3a2b(a + b)

Sol. a(a + b)3 – 3a2b(a + b)

= a(a + b) {(a + b)2 – 3ab}

= a(a + b) {a2 + b2 + 2ab – 3ab}

= a(a + b) (a2 – ab + b2)

3. x(x3 – y3) + 3xy(x – y)

Sol. x(x3 – y3) + 3xy(x – y)

= x(x – y) (x2 + xy + y2) + 3xy(x – y)

= x(x – y) (x2 + xy + y2 + 3y)

= x(x – y) (x2 + xy + y2 + 3y)

4. a2x2 + (ax2 + 1)x + a

Sol. a2x2 + (ax2 + 1)x + a

= a2x2 + a + (ax2 + 1)x

= a(ax2 + 1) + x(ax2 + 1)

= (ax2 + 1) (a + x) = (x + a) (ax2 + 1)

5. x2 + y – xy – x

Sol. x2 + y – xy – x

= x2 – x – xy + y

= x(x – 1) – y(x – 1)

= (x – 1) (x – y)

6. x3 – 2x2y + 3xy2 – 6y3

Sol. x3 – 2x2y + 3xy2 – 6y3

= x2(x – 2y) + 3y2(x – 2y)

= (x – 2y) (x2 + 3y2)

7. 6ab – b2 + 12ac – 2bc

Sol. 6ab – b2 + 12ac – 2bc

= 6ab + 12ac – b2 – 2bc

= 6a(b + 2c) – b(b + 2c)

= (b + 2c) (6a – b)

8. x(x – 2) (x – 4) + 4x – 8

Sol. x(x – 2) (x – 4) + 4x – 8

= x(x – 2) (x – 4) + 4(x – 2)

= (x – 2) [x(x – 4) + 4]

= (x – 2) (x2 – 4x + 4)

= (x – 2) [(x)2 – 2 × x × 2 + (2)2]

= (x – 2) (x – 2)2 = (x – 2)3

9. (a – b + c)2 + (b – c + a)2 + 2(a – b + c)

(b – c + a)

Sol. (a – b + c)2 + (b – c + a)2 + 2(a – b + c)

(b – c + a)

{_ a2 + b2 + 2ab = (a + b)2}

= [a – b + c + b – c + a]2

= (2a)2 = 4a2

10. a2 + 2ab + b2 – c2

Sol. a2 + 2ab + b2 – c2

= (a2 + 2ab + b2) – c2

= (a + b)2 – (c)2

{_ a2 – b2 = (a + b) (a – b)}

= (a + b + c) (a + b – c)

11. a2 + 4b2 – 4ab – 4c2

Sol. a2 + 4b2 – 4ab – 4c2

= (a)2 + (2b)2 – 2 × a × 2b – (2c)2

= (a – 2b)2 – (2c)2

(a – 2b + 2c) (a – 2b – 2c)

22

222

)()(

)(2

bababa

bababa

12. x2 – y2 – 4xz + 4z2

Sol. x2 – y2 – 4xz + 4z2

= x2 – 4xz + 4z2 – y2

FACTORIZATION OF ALGEBRAIC EXPRESSIONS5



= (x)2 – 2 × x × 2z + (2z)2 – (y)2

= (x – 2z)2 – (y)2

= (x – 2z + y) (x – 2z – y)

= (x +y – 2z) (x – y – 2z)

13. 2x2 – 6

5x +

12

1

Sol. 2x2 – 6

5x +

12

1

2x2 – 2

1x –

3

1x +

12

1

x

2

12x –

6

1

2

12x

3

1

2

1

6

5

3

1

2

1

6

1

6

1

12

2

12

12

2

12x

6

1x

14. x2 + 35

12x +

35

1

Sol. x2 + 35

12x +

35

1

7

1

5

1

35

12

7

1

5

1

35

1

= x2 + 5

1x +

7

1x +

35

1

= x

5

1x +

7

1

5

1x

=

5

1x

7

1x

15. 21x2 – 2x + 21

1

Sol. 21x2 – 2x + 21

1

= 221x – 2 × 21x × 21

1 +

2

21

1

=

2

21

121

x

16. Give possible expression for the length and

breadth of the rectangle having 35y2 + 13y

– 12 as its area.

Sol. Area of a rectangle = 35y2 + 13y – 12

= 35y2 + 28y – 15y – 12

152813

)15(28420

420)12(35

= 7y(5y + 4) – 3(5y + 4)

= (5y + 4) (7y – 3)

(i) If length = 5y + 4, then breadth = 7y – 3

(ii) and if length = 7y – 3, then length = 5y + 4

17. What are the possible expressions for the

dimensions of the cuboid whose volume is

3x2 – 12x.

Sol. Volume 3x2 – 12x

= 3x(x – 4)

Factors are 3, x, and x – 4

Now, if length = 3, breadth = x and height

= x – 4

if length = 3, breadth = x – 4, height = x

if length = x, breadth = 3, height = x – 4

if length = x, breadth = x – 4, height = 3

if length = x – 4, breadth = 3, height = x

if length = x – 4, breadth = x, height = 3

18.

2

2 1

xx – 4

xx

1 + 6



Sol.

2

2 1

xx – 4

xx

1 + 6

212

2

xx – 4

xx

1 + 4

2

1

xx – 2 × 2

xx

1 + (2)2

=

2

21

x

x {_ a2 – 2ab + b2 = (a – b)2}

19. (x + 2) (x2 + 25) – 10x2 – 20x

Sol. (x + 2) (x2 + 25) – 10x2 – 20x

= (x + 2) (x2 + 25) – 10x(x + 2)

= (x + 2) [x2 + 25 – 10x]

= (x + 2) [(x)2 – 2 × x × 5 + (5)2]

= (x + 2) (x – 5)2

20. 2a2 + 2 6 ab + 3b2

Sol. 2a2 + 2 6 ab + 3b2

= ( 2 a)2 + 2 × 2 a × 3 b + ( 3 b)2

= ( 2 a + 3 b)2

21. a2 + b2 + 2(ab + bc + ca)

Sol. a2 + b2 + 2(ab + bc + ca)

= a2 + b2 + 2ab + 2bc + 2ca

= (a + b)2 + 2c(b + a)

= (a + b)2 + 2c(a + b)

= (a + b) (a + b + 2c)

22. 4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2

Sol. 4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2

= [2(x – y)]2 + 2 × 2(x – y) × 3(x + y) +

[3(x + y]2

{_ a2 + b2 + 2abc = (a + b)2}

= [2(x – y) + 3(x + y)]2

= (2x – 2y + 3x + 3y)2 = (5x + y)2

23. a2 – b2 + 2bc – c2

Sol. a2 – b2 + 2bc – c2

= a2 – (b2 – 2bc + c2)

{_ a2 + b2 – 2abc = (a – b)2}

= a2 – (b – c)2

= (a)2 – (b – c)2

{_ a2 – b2 = (a + b) (a – b)}

= (a + b – c) (a – b + c)

24. xy9 – yx9

Sol. xy9 – yx9 = xy(y8 – x8)

= –xy(x8 – y8)

= –xy[(x4)2 – (y4)2]

= –xy(x4 + y4) (x4 – y4)

{_ a2 – b2 = (a + b) (a – b)}

= –xy (x4 + y4) {(x2)2 – (y2)2}

= –xy(x4 + y4) (x2 + y2) (x2 – y2)

= –xy(x4 + y4) (x2 + y2) (x + y) (x – y)

= –xy(x – y) (x + y) (x2 + y2) (x4 + y4)

25. x4 + x2y2 + y4

Sol. x4 + x2y2 + y4 = (x2)2 + 2x2y2 + y4 – x2y2

(Adding and subtracting x2y2)

= (x2 + y2)2 – (xy)2

{_ a2 – b2 = (a + b) (a – b)}

= (x2 + y2 + xy) (x2 + y2 – xy)

= (x2 + xy + y2) (x2 – xy + y2)

26. x2 + 6 2 x + 10

Sol. x2 + 6 2 x + 10

= x2 + 5 2 x + 2 x + 10

22510

22526

= x(x + 5 2 ) + 2 (x + 5 2 )

= (x + 5 2 ) (x + 2 )

27. x2 – 2 2 x – 30

Sol. x2 – 2 2 x – 30

= x2 – 5 2 x + 3 2 x – 30

= x(x – 5 2 ) + 3 2 (x – 5 2 x)

= (x – 5 2 ) (x + 3 2 )

28. x2 – 3 x – 6



Sol. x2 – 3 x – 6

= x2 – 2 3 x + 3 x – 6

= x(x – 2 3 ) + 3 (x – 2 3 )

3323

3326

= (x – 2 3 ) (x + 3 )

29. x2 + 5 5 x + 30

Sol. x2 + 5 5 x + 30

= x2 + 3 5 x + 2 5 + 30

525330

525330 xx

= x(x + 3 5 ) + 2 5 (x + 3 5 )

= (x + 3 5 ) (x + 2 5 )

30. x2 + 2 3 x – 24

Sol. x2 + 2 3 x – 24

323432

)32(3424

x2 + 4 3 x – 2 3 x – 24

x(x + 4 3 ) – 2 3 (x + 4 3 )

= (x + 4 3 ) (x – 2 3 )

31. 5 5 x2 + 20x + 3 5

Sol. 5 5 x2 + 20x + 3 5

= 5 5 x2 + 5x + 15x + 3 5

51520

51575

755355

= 5 x(5x + 5 ) + 3(5x + 5 )

= (5x + 5 ) ( 5 x + 3)

32. 2x2 + 3 5 x + 5

Sol. 2x2 + 3 5 x + 5

55253

55210

1052

= 2x2 + 2 5 x + 5 x + 5

= 2x(x + 5 ) + 5 (x + 5 )

= (x + 5 ) (2x + 5 )

33. 9(2a – b)2 – 4(2a – b) – 13

Sol. 9(2a – b)2 – 4(2a – b) – 13

Let 2a – b = x, then

9x2 – 4x – 13

9134

913117

117)13(9

9x2 + 9x – 13x – 13

9x(x + 1) – 13(x + 1)

(x + 1) (9x – 13)

(2a – b + 1) {9 (2a – b) – 13}

(2a – b + 1) (18a – 9b – 13)

34. 7(x – 2y)2 – 25(x – 2y) + 12

Sol. 7(x – 2y)2 – 25(x – 2y) + 12

Let x – 2y = a

7a2 – 25a + 12

42125

)4(2184

84127

= 7a2 – 21a – 4a + 12

= 7a(a – 3) – 4(a – 3)

= (a – 3) (7a – 4)

= (x – 2y – 3) (7x – 14y – 4)

35. 2(x + y)2 – 9(x + y) – 5

Sol. 2(x + y)2 – 9(x + y) – 5



Let x + y = a, then

2a2 – 9a – 5

1109

11010

10)5(2

= 2a2 – 10a + a – 5

= 2a(a – 5) + 1(a – 5)

= (a – 5) (2a + 1)

= (x + y – 5) (2x + 2y + 1)

EXERCISE 5.2

Factorize each of the following expressions:

1. p3 + 27

Sol. We know that a3 + b3 = (a + b) (a2 – ab +

b2)

a3 – b3 = (a – b) (a2 + ab + b2)

p3 + 27 = (p)3 + (3)3

= (p + 3) (p2 – p × 3 + 32)

= (p + 3) (p2 – 3p + 9)

2. y3 + 125

Sol. y3 + 125 = (p)3 + (5)3

= (p + 5) (p2 – 5y + 52)

= (p + 5) (p2 – 5y + 25)

3. 1 – 27a3

Sol. 1 – 27a3 = (1)3 – (3a)3

= (1 – 3a) [12 + 1 × 3a + (3a)2]

= (1 – 3a) (1 + 3a + 9a2)

4. 8x3y3 + 27a3

Sol. 8x3y3 + 27a3

= (2xy + 3a) [(2xy)2 – 2xy × 3a + (3a)2]

= (2xy + 3a) (4x2y2 – 6xya + 9a2)

5. 64a3 – b3

Sol. 64a3 – b3 = (4a)3 – (b)3

= (4a – b) [(4a)2 + 4a × b + (b)2]

= (4a – b) (16a2 + 4ab + b2)

6.216

3x

– 8y3

Sol.216

3x

– 8y3 =

3

6

x

– (2y)3

=

yx

26

2

2

)2(266

yyxx

=

yx

26

2

2

4336

yxyx

7. 10x4y – 10xy4

Sol. 10x4y – 10xy4 = 10xy(x3 – y3)

= 10xy(x – y) (x2 + xy + y2)

8. 54x6y + 2x3y4

Sol. 54x6y + 2x3y4 = 2x3y(27x3 + y3)

= 2x3y[(3x)3 + (y)3]

= 2x3y(3x + y) [(3x)2 – 3x × y + y2]

= 2x3y(3x + y) (9x2 – 3xy + y2)

9. 32a3 + 108b3

Sol. 32a3 + 108b3

= 4(8a3 + 27b3) = 4[(2a)3 + (3b)3]

= 4(2a + 3b) [(2a)2 – 2a × 3b + (3b)2]

= 4(2a + 3b) (4a2 – 6ab + 9b2)

10. (a – 2b)3 – 512b3

Sol. (a – 2b)3 – 512b3

= (a – 2b)3 – (8b)3

= (a – 2b – 8b) [(a – 2b)2 + (a – 2b) × 8b

+ (8b)2]

= (a – 10b) [a2 + 4b2 – 4ab + 8ab – 16b2 +

64b2]

= (a – 10b) (a2 + 4ab + 52b2)

11. 8x2y3 – x5

Sol. 8x2y3 – x5 = x2(8y3 – x3)

= x2[(2y)3 – (x)3]

= x2[(2y – x) (2y)2 + 2y × x + (x)2]

= x2(2y – x) (4y2 + 2xy + x2)

12. 1029 – 3x3

Sol. 1029 – 3x3 = 3(343 – x3)

= 3[(7)3 – (x)3]

= 3(7 – x) (49x + 7x + x2)

13. x3y3 + 1

Sol. x3y3 + 1 = (xy)3 + (1)3

= (xy + 1) [(xy)2 – xy × 1 + (1)2]

= (xy + 1) (x2y2 – xy + 1)

= (xy + 1) (x2y2 – xy + 1)



14. x4y4 – xy

Sol. x4y4 – xy = xy(x3y3 – 1)

= xy[(xy)3 – (1)3]

= xy(xy – 1) [x2y2 + 2xy + 1]

15. a3 + b3 + a + b

Sol. a3 + b3 + a + b

= (a + b) (a2 – ab + b2) + 1(a + b)

= (a + b) (a2 – ab + b2 + 1)

16. Simplify:

(i)127127127173173173

127127127173173173

(ii)555555155155155

555555155155155

(iii)2.02.02.02.12.12.1

2.02.02.02.12.12.1

Sol. (i) 127127127173173173

127127127173173173

Let 173 = a and 127 = b, then

= 22

33

baba

ba

= 22

22)()(

baba

bababa

= a + b = 173 + 127 = 300

(ii)555555155155155

555555155155155

Let 155 = a and 55 = b, then

= 22

33

baba

ba

= 22

22)()(

baba

bababa

= a – b = 155 – 55 = 100

(iii)2.02.02.02.12.12.1

2.02.02.02.12.12.1

Let 1.2 = a and 0.2 = b, then

22

33

baba

ba

= 22

22)()(

baba

bababa

= a – b

= 1.2 – 0.2 = 1.0 = 1

17. (a + b)3 – 8(a – b)3

Sol. (a + b)3 – 8(a – b)3

= (a + b)3 – (2a – 2b)3

= (a + b – 2a + 2b) [(a + b)2 + (a + b) (2a

– 2b) + (2a – 2b)2)]

= (3b – a) [a2 + b2 + 2ab + 2a2 – 2ab +

2ab – 2b2 + 4a2 – 8ab + 4b2]

= (3b – a) [7a2 – 6ab + 3b2]

18. (x + 2)3 + (x – 2)3

Sol. (x + 2)3 + (x – 2)3

= (x + 2 + x – 2) [(x + 2)2 – (x + 2) (x – 2)

+ (x – 2)2]

= 2x [x2 + 4x + 4 – (x2 + 2x – 2x – 4) + x2

– 4x + 4]

= 2x[x2 + 4x + 4 – x2 – 2x + 2x + 4 + x2 –

4x + 4]

= 2x[x2 + 12]

19. x6 + y6

Sol. x6 + y6 = (x2)3 + (y2)3

= (x2 + y2) [x4 – x2y2 + y4]

20. a12 + b12

Sol. a12 + b12 = (a4)3 + (b4)3

= (a4 + b4) [(a4)2 – a4b4 + (b4)2]

= (a4 + b4) (a8 – a4b4 + b8)

21. x3 + 6x2 + 12x + 16

Sol. x3 + 6x2 + 12x + 16

= (x)3 + 3.x2.2 + 3.x.4 + (2)3 + 8

{_ a3 + 3a2b + 3ab2 + b3 = (a + b)3}

= (x + 2)3 + 8 = (x + 2)3 + (2)3

= (x + 2 + 2) [(x + 2)2 – (x + 2) × 2 + (2)2]

{_ a3 + b2 = (a + b) (a2 – ab + b2}

= (x + 4) (x2 + 4x + 4 – 2x – 4 + 4)

= (x + 4) (x2 + 2x + 4)

22. a3 – 3

1

a – 2a +

a

2

Sol. a3 – 3

1

a – 2a +

a

2

=

a

a1

2

2 11

aa – 2

a

a1

=

a

a1

21

12

2

aa



=

a

a1

2

2 11

aa

=

a

a1

11

2

2

aa

23. a3 + 3a2b + 3ab2 + b3 – 8

Sol. a3 + 3a2b + 3ab2 + b3 – 8

= (a + b)3 – (2)3

= (a + b – 2) [(a + b)2 + (a + b) × 2 + (2)2]

= (a + b – 2) (a2 + b2 + 2ab + 2a + 2b + 4)

= (a + b – 2) (a2 + b2 + 2ab + 2(a + b) + 4]

= (a + b – 2) [(a + b)2 + 2(a + b) + 4]

24. 8a3 – b3 – 4ax + 2bx

Sol. 8a3 – b3 – 4ax + 2bx

(2a)3 – (b)3 – 2x(2a – b)

= (2a – b) [(2a)2 + 2a × b + (b)2] – 2x(2a – b)

= (2a – b) [4a2 + 2ab + b2] – 2x(2a – b)

= (2a – b) [4a2 + 2ab + b2 – 2x]

EXERCISE 5.3

Factorize:

1. 64a3 + 125b3 + 240a2b + 300ab2

Sol. 64a3 + 125b3 + 240a2b + 300ab2

= (4a)3 + (5b)3 + 3 × (4a)2 × 5b + 3(4a) +

(5b)2

= (4a + 5b)3 = (4a + 5b) (4a + 5b) (4a +

5b)

2. 125x3 – 27y3 – 225x2y + 135xy2

Sol. 125x3 – 27y3 – 225x2y + 135xy2

= (5x)3 – (3y)3 – 3 × (5x)2 × (3y) + 3 × 5x

× (3y)2

= (5x – 3y)3 = (5x – 3y) (5x – 3y) (5x – 3y)

3.27

8x3 + 1 +

3

4x2 + 2x

Sol.27

8x3 + 1 +

3

4x2 + 2x

=

3

3

2

x + (1)3 + 3 ×

2

3

2

x × 1 + 3 ×

3

2x × (1)2

=

3

13

2

x =

13

2x

13

2x

13

2x

4. 8x3 + 27y3 + 36x2y + 54xy2

Sol. 8x3 + 27y3 + 36x2y + 54xy2

= (2x)3 + (3y)3 + 3 × (2x)2 × 3y + 3 × 2x ×

(3y)2

= (2x + 3y)3 = (2x + 3y) (2x + 3y) (2x + 3y)

5. a3 – 3a2b + 3ab2 – b3 + 8

Sol. a3 – 3a2b + 3ab2 – b3 + 8

= (a – b)3 + (2)3

= (a – b + 2) [(a – b)2 – (a – b) × 2 + (2)2]

= (a – b + 2) (a2 + b2 – 2ab – 2a + 2b + 4)

6. x3 + 8y3 + 6x2y + 12xy2

Sol. x3 + 8y3 + 6x2y + 12xy2

= (x)3 + (2y)3 + 3 × x2 × 2y + 3 × x × (2y)2

= (x + 2y)3 = (x + 2y) (x + 2y) (x + 2y)

7. 8x3 + y3 + 12x2y + 6xy2

Sol. 8x3 + y3 + 12x2y + 6xy2

= (2x)3 + (y)3 + 3 × (2x)2 × y + 3 × 2x × y2

= (2x + y)3 = (2x + y) (2x + y) (2x + y)

8. 8a3 + 27b3 + 36a2b + 54ab2

Sol. 8a3 + 27b3 + 36a2b + 54ab2

= (2a)3 + (3b)3 + 3 × (2a)2 × 3b + 3 × 2a

× (3b)2

= (2a + 3b)3 = (2a + 3b) (2a + 3b) (2a + 3b)

9. 8a3 – 27b3 – 36a2b + 54ab2

Sol. 8a3 – 27b3 – 36a2b + 54ab2

= (2a)3 – (3b)3 – 3 × (2a)2 × 3b + 3 × 2a ×

(3b)2

= (2a – 3b)3 = (2a – 3b) (2a – 3b) (2a – 3b)

10. x3 – 12x(x – 4) – 64

Sol. x3 – 12x(x – 4) – 64

= x3 – 12x2 + 48x – 64

= (x)3 – 3 × x2 × 4 + 3 × x × (4)2 – (4)3

= (x – 4)3 = (x – 4) (x – 4) (x – 4)

11. a3x3 – 3a2bx2 + 3ab2x – b3

Sol. a3x3 – 3a2bx2 + 3ab2x – b3

= (ax)3 – 3 × (ax)2 × b + 3 × ax × (b)2 – (b)3

= (ax – b)3 = (ax – b) (ax – b) (ax – b)



EXERCISE 5.4

Factorize each of the following expressions:

1. a3 + 8b3 + 64c3 – 24abc

Sol. We know that a3 + b3 + c3 – 3abc = (a + b

+ c) (a2 + b2 + c2 – ab – bc – ca)

a3 + 8b3 + 64c3 – 24abc

= (a)3 + (2b)3 + (4c)3 – 3 × a × 2b × 4c

= (a + 2b + 4c) [(a)2 + (2b)2 + (4c)2 – a ×

2b – 2b × 4c – 4c × a]

= (a + 2b + 4c) (a2 + 4b2 + 16c2 – 2ab –

8bc – 4ca)

2. x3 – 8y3 + 27z3 + 18xyz

Sol. x3 – 8y3 + 27z3 + 18xyz

= (x)3 + (–2y)3 + (3z)3 – 3 × x × (–2y) (3z)

= (x – 2y + 3z) (x2 + 4y2 + 9z2 + 2xy + 6yz

– 3zx)

3. 27x3 – y3 – z3 – 9xyz [NCERT]

Sol. 27x3 – y3 – z3 – 9xyz

= (3x)3 + (–y)3 + (–z)3 – 3 × 3x × (–y) (–z)

= (3x – y – z) [(3x)2 + (–y)2 + (–z)2 – 3x ×

(–y) – (–y) (–z) – (–z × 3x)]

= (3x – y – z) (9x2 + y2 + z2 + 3xy – yz +

3zx)

4.27

1x3 – y3 + 125z3 + 5xyz

Sol.27

1x3 – y3 + 125z3 + 5xyz

=

3

3

1

x + (–y)3 + (5z)3 – 3 × 3

x × (–y) × 5z

=

zyx 53

1

2

3

1x + (–y)2 + (5z)2 –

3

1x × (–y) – (–y) × 5z) – 5z ×

x

3

1

=

zyx 53

1 2

9

1x + y2 + 25z2 +

3

1xy

+ 5yz –

zx3

5

5. 8x3 + 27y3 – 216z3 + 108xyz

Sol. 8x3 + 27y3 – 216z3 + 108xyz

= (2x)3 + (3y)3 + (6z)3 – 3 × (2x) (3y) (–6z)

= (2x + 3y – 6z) [(2x)2 + (3y)2 + (–6z)2 –

2x × 3y – 3y × (–6z) – (–6z) × 2x]

= (2x + 3y – 6z) (4x2 + 9y2 + 36z2 – 6xy +

18yz + 12zx)

6. 125 + 8x3 – 27y3 + 90xy

Sol. 125 + 8x3 – 27y3 + 90xy

= (5)3 + (2x)3 + (–3y)3 – [3 × 5 × 2x × (–3y)]

= (5 + 2x – 3y) [(5)2 + (2x)2 + (–3y)2 – 5 ×

2x – 2x (–3y) – (–3y) × 5]

= (5 + 2x – 3y) (25 + 4x2 + 9y2 – 10x + 6xy

+ 15y)

7. 8x3 – 125y3 + 180xy + 216

Sol. 8x3 – 125y3 + 180xy + 216

= (2x)3 + (–5y)3 + (6)3 – 3 × 2x (–5y) × 6

= (2x – 5y + 6) [(2x)2 + (–5y)2 + (6)2 – 2x

× (–5y) – (–5y) × 6 – 6 × 2x]

= (2x – 5y + 6) (4x2 + 25y2 + 36 + 10xy +

30y – 12x

8. Multiply:

(i) x2 + y2 + z2 – xy + xz + yz by x + y – z

(ii) x2 + 4y2 + z2 + 2xy + xz – 2yz by x – 2y – z

(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3

(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x

– 5y + 4

Sol. (i) (x2 + y2 + z2 – xy + yz + zx)

by (x + y – z)

= x3 + y3 – z3 + 3xyz

(ii) (x2 + 4y2 + z2 + 2xy + xz – 2yz)

by (x – 2y – z)

= (x – 2y – z) [x2 + (–2y)2 + (–z)2 – x × (–

2y) – (–2y) (z) – (–z) (x)]

= x3 + (–2y)3 + (–z)3 – 3x (–2y) (–z)

= x3 – 8y3 – z3 – 6xyz

(iii) x2 + 4y2 + 2xy – 3x + 6y + 9 by x – 2y + 3

= (x – 2y + 3) (x2 + 4y2 + 9 + 2xy + 6y –

3x)

= (x)3 + (–2y)3 + (3)3 – 3 × x × (–2y) × 3

= x3 – 8y3 + 27 + 18xy

(iv) 9x2 + 25y2 + 15xy + 12x – 20y + 16 by 3x

– 5y + 4



= (3x – 5y + 4) [(3x)2 + (–5y)2 + (4)2 – 3x

× (–5y) (–5y × 4) – (4 × 3x)]

= (3x)3 + (–5y)3 + (4)3 – 3 × 3x (–5y) × 4

= 27x3 – 125y3 + 64 + 180xy

9. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

Sol. (3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

_ 3x – 2y + 2y – 4z + 4z – 3x = 0

(3x – 2y)3 + (2y – 4z)3 + (4z – 3x)3

= 3(3x – 2y) (2y – 4z) (4z – 3x)

{_ x3 + y3 + z3 = 3xyz if x + y + z = 0}

10. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

Sol. (2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

_ 2x – 3y + 4z – 2x + 3y – 4z = 0

(2x – 3y)3 + (4z – 2x)3 + (3y – 4z)3

= (2x – 3y) (4z – 2x) (3y – 4z)

{_ x3 + y3 + z3 = 3xyz if x + y + z = 0}

11.

3

32

z

yx

+

3

3

2

3

zyx

+

3

3

4

36

5

zyx

Sol.

3

32

z

yx

+

3

3

2

3

zyx

+

3

3

4

36

5

zyx

_

2

x + y +

3

z +

3

x –

3

2y + z +

6

5x –

3

y

– 3

4z

= 2

x +

3

x –

6

5x + y –

3

2y –

3

y +

3

z + z

– 3

4z

= 6

523 xxx +

3

23 yyy +

3

43 zzz

= 0 + 0 + 0 = 0

_

3

32

z

yx

+

3

3

2

3

zyx

+

3

3

4

36

5

zyx

=

32

zy

x

zyx

3

2

3

3

4

36

5 zyx

12. (a – 3b)3 + (3b – c)3 + (c – a)3

Sol. (a – 3b)3 + (3b – c)3 + (c – a)3

_ a – 3b + 3b – c + c – a = 0

(a – 3b)3 + (3b – c)3 + (c – a)3

= 3(a – 3b) (3b – c) (c – a)

{_ a3 + b3 + c3 = 3abc if a + b + c = 0}

13. 2 2 a3 + 3 3 b3 + c3 – 3 6 abc

Sol. 2 2 a3 + 3 3 b3 + c3 – 3 6 abc

= ( 2 a)3 + ( 3 b)3 + (c)3 – 3 × 2 a ×

3 b × c

= ( 2 a + 3 b + c) [( 2 a)2 + ( 3 b)2 +

c2 – 2 a × 3 b – 3 b × c – c × 2 a]

= ( 2 a + 3 b + c) (2a2 + 3b2 + c2 –

6 ab – 3 bc – 2 ca)

14. 3 3 a3 – b3 – 5 5 c3 – 3 15 abc

Sol. 3 3 a3 – b3 – 5 5 c3 – 3 15 abc

( 3 a)3 + (–b)3 + (– 5 c)3 – 3 3 a × (–b)

× (– 5 c)

= ( 3 a – b – 5 c) [( 3 a)2 + (–b)2 +

(– 5 c)2 – 3 a × (–b) – (–b) (– 5 c) –

(– 5 c) × 3 a



= ( 3 a – b – 5 c) (3a2 + b2 + 5c2 +

3 ab – 5 bc + 15 ca)

15. 2 2 a3 + 16 2 b3 + c3 – 12abc

Sol. (2 2 a)3 + (2 2 b)3 + (c)3 – 3 × 2 a ×

2 2 b × c

= ( 2 a + 2 2 b + c) [( 2 a)2 + (2 2 b)2

+ c2 – 2 a × 2 2 b – 2 2 b × c – c ×

2 a

= ( 2 a + 2 2 b + c) (2a2 + 8b2 + c2 –

4ab – 2 2 bc – 2 ca)

16. Find the value of x3 + y3 – 12xy + 64, when

x + y = –4

Sol. x3 + y3 – 12xy + 64

x + y = –4

Cubing both sides,

x3 + y3 + 3xy(x + y) = –64

Substitute the value of (x + y)

x3 + y3 + 3xy × (–4) = –64

x3 + y3 – 12xy + 64 = 0


(VSAQs)

1. If a + b + c = 0, then write the value of a3

+ b3 + c3.

Sol. _ a + b + c = 0,

Then a3 + b3 + c3 = 3abc

2. If a2 + b2 + c2 = 20 and a + b + c = 0, find

ab + bc + ca.

Sol. a2 + b2 + c2 = 20, a + b + c = 0

(a + b + c)2 = 0

a2 + b2 + c2 + 2(ab + bc + ca) = 0

20 + 2(ab + bc + ca) = 0

2(ab + bc + ca) = –20

ab + bc + ca = 2

20 = –10

3. If a + b + c = 9 and ab + bc + ca = 40,

find a2 + b2 + c2.

Sol. a + b + c = 9, ab + bc + ca = 40


(a + b + c)2 = (9)2

a2 + b2 + c2 + 2(ab + bc + ca) = 81

a2 + b2 + c2 + 2 × 40 = 81

a2 + b2 + c2 + 80 = 81

a2 + b2 + c2 = 81 – 80 = 1

4. If a2 + b2 + c2 = 250 and ab + bc + ca = 3,

find a + b + c.

Sol. a2 + b2 + c2 = 250, ab + bc + ca = 3

(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

= 250 + 2 × 3 = 250 + 6 = 256

= (+16)2

a + b + c = +16

5. Write the value of 253 – 753 + 503.

Sol. 253 – 753 + 503

Let a = 25, b = –75 and c = 50

_ a + b + c = 25 – 75 + 50 = 0

a3 + b3 + c3 = 3abc

253 + (–75)3 + 503 = 3 × 25 × (–75) × 50

= –281250

6. Write the value of 483 – 303 – 183.

Sol. 483 – 303 – 183

Let a = 48, b = –30, c = –18

_ a + b + c = 48 – 30 – 18 = 0

a3 + b3 + c3 = 3abc

483 – 303 – 183 = 3 × 48 × (–30) (–18)

= 77760

7. Write the value of

3

2

1

+

3

3

1

–

3

6

5

.

Sol.

3

2

1

+

3

3

1

–

3

6

5

Let a = 2

1, b =

3

1 and c =

6

5

_ a + b + c = 2

1 +

3

1 –

6

5

= 6

523 =

6

0 = 0



a3 + b3 + c3 = 3abc

3

2

1

+

3

3

1

–

3

6

5

= 3 × 2

1 ×

3

1 ×

6

5 =

12

5

8. Write the value of 303 + 203 – 503.

Sol. 303 + 203 – 503

Let a = 30, b = 20, c = –50

_ a + b + c = 30 + 20 – 50 = 50 – 50 = 0

a3 + b3 + c3 = 3abc

303 + 203 – 503 = 3 × 30 × 20 × (–50)

= 90000

9. Factorize: x4 + x2 + 25.

Sol. x4 + x2 + 25

(x2)2 + (5)2 + 2x2 × 5 – 2x2 × 5 + x2

(x2)2 + (5)2 + 10x2 – 10x2 + x2

= (x2)2 + (5)2 + 10x2 – 9x2

= (x2 + 5)2 – (3x)2

{_ a2 – b2 = (a + b) (a – b)}

= (x2 + 5 – 3x) (x2 + 5 + 3x)

= (x2 – 3x + 5) (x2 + 3x + 5)

10. Factorize: x2 – 1 – 2a – a2.

Sol. x2 – 1 – 2a – a2

= x2 – (1 + 2a + a2) = (x)2 – (1 + a)2

{_ a2 – b2 = (a + b) (a – b)}

= (x + 1 + a) (x – 1 – a)

= (x + a + 1) (x – a – 1)



following:

1. The factors of x3 – x2y – xy2 + y3 are

(a) (x + y) (x2 – xy + y2)

(b) (x + y) (x2 + xy + y2)

(c) (x + y)2 (x – y)

(d) (x – y)2 (x + y)

Sol. x3 – x2y – xy2 + y3

= x3 + y3 – x2y – xy2

= (x + y) (x2 – xy + y2) – xy(x + y)

= (x + y) (x2 – xy + y2 – xy)

= (x + y) (x2 – 2xy + y2) = (x + y) (x – y)2

(d)

2. The factors of x3 – 1 + y3 + 3xy are

(a) (x – 1 + y) (x2 + 1 + y2 + x + y – xy)

(b) (x + y + 1) (x2 + y2 + 1 – xy – x – y)

(c) (x – 1 + y) (x2 – 1 – y2 + x + y + xy)

(d) 3(x + y – 1) (x2 + y2 – 1)

Sol. x3 – 1 + y3 + 3xy

= (x)3 + (–1)3 + (y)3 – 3 × x × (–1) × y

= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)

= (x – 1 + y) (x2 + 1 + y2 + x + y – xy)

(a)

3. The factors of 8a3 + b3 – 6ab + 1 are

(a) (2a + b – 1) (4a2 + b2 + 1 – 3ab – 2a)

(b) (2a – b + 1) (4a2 + b2 – 4ab + 1 – 2a + b)

(c) (2a + b + 1) (4a2 + b2 + 1 – 2ab – b – 2a)

(d) (2a – 1 + b) (4a2 + 1 – 4a – b – 2ab)

Sol. 8a3 + b3 – 6ab + 1

= (2a)3 + (b)3 + (1)3 – 3 × 2a × b × 1

= (2a + b + 1) [(2a)2 + b2 + 1 – 2a × b –

b × 1 – 1 × 2a]

= (2a + b + 1) (4a2 + b2 + 1 – 2ab – b –

2a) (c)

4. (x + y)3 – (x – y)3 can be factorized as

(a) 2y(3x2 + y2) (b) 2x(3x2 + y2)

(c) 2y(3y2 + x2) (d) 2x(x2 + 3y2)

Sol. (x + y)3 – (x – y)3

= (x + y – x + y) [(x + y)2 + (x + y) (x – y)

+ (x – y)2]

= 2y(x2 + y2 + 2xy + x2 – y2 + x2 + y2 – 2xy)

= 2y(3x2 + y2) (a)

5. The expression (a – b)3 + (b – c)3 + (c –

a)3 can be factorized as

(a) (a – b) (b – c) (c – a)

(b) 3(a – b) (b – c) (c – a)

(c) –3(a – b) (b – c) (c – a)

(d) (a + b + c) (a2 + b2 + c2 – ab – bc – ca)

Sol. (a – b)3 + (b – c)3 + (c – a)3

Let a – b = x, b – a = y, c – a = z

x3 + y3 + z3

x + y + z = a – b + b – c + c – a = 0



x3 + y3 + z3 = 3xyz

(a – b)3 + (b – a)3 + (c – a)3

= 3(a – b) (b – c) (c – a) (b)

6. The value of 09.069.0)3.2(

027.0)3.2(2

3

is

(a) 2 (b) 3 (c) 2.327 (d) 2.273

Sol.09.069.0)3.2(

027.0)3.2(2

3

{_ a3 – b3 = (a – b) (a2 + ab + b2)}

= 22

33

)3.0()3.0()3.2()3.2(

)3.0()3.2(

= 22

22

)3.0()3.0()3.2()3.2(

})3.0()3.0()3.2()3.2{()}3.0()3.2{(

= (2.3) – (0.3) = 2.3 – 0.3 = 2 (a)

7. The value of 22

33

)007.0(007.0013.0)013.0(

)007.0()013.0(

is

(a) 0.006 (b) 0.02 (c) 0.0091 (d) 0.00185

Sol. 22

33

)007.0(007.0013.0)013.0(

)007.0()013.0(

{_ a3 + b3 = (a + b) (a2 – ab + b2)}

= 22

22

)007.0(007.0013.0)013.0(

])007.0(007.0013.0)013.0[()007.0013.0(

= 0.013 + 0.007 = 0.020 = 0.02 (b)

8. The factors of a2 – 1 – 2x – x2 are

(a) (a – x + 1) (a – x – 1) (b) (a + x – 1) (a – x + 1)

(c) (a + x + 1) (a – x – 1) (d) none of these

Sol. a2 – 1 – 2x – x2

a2 – (1 + 2x + x2) = (a)2 – (1 + x)2

= (a + 1 + x) (a – 1 – x) (c)

9. The factors of x4 + x2 + 25 are

(a) (x2 + 3x + 5) (x2 – 3x + 5) (b) (x2 + 3x + 5) (x2 + 3x – 5)

(c) (x2 + x + 5) (x2 – x + 5) (d) none of these

Sol. x4 + x2 + 25 = x4 + 25 + x2

= (x2)2 + (5)2 + 2 × x2 × 5 – 9x2

= (x2 + 5)2 – (3x)2

= (x2 + 5 + 3x) (x2 + 5 – 3x)

= (x2 + 3x + 5) (x2 – 3x + 5) (a)



10. The factors of x2 + 4y2 + 4y – 4xy – 2x – 8 are

(a) (x – 2y – 4) (x – 2y + 2) (b) (x – y + 2) (x – 4y – 4)

(c) (x + 2y – 4) (x + 2y + 2) (d) none of these

Sol. x2 + 4y2 + 4y – 4xy – 2x – 8

x2 + 4y2 + 4y – 4xy – 2x – 8

= (x)2 + (2y)2 – 2 × x × 2y + 4y – 2x – 8

= (x – 2y)2 – (2x – 4y) – 8

= (x – 2y)2 – 2 (x – 2y) – 8

Let x – 2y = a, then

a2 – 2a – 8 = a2 – 4a + 2a – 8

= a(a – 4) + 2(a – 4)

= (a – 4) (a + 2)

= (x2 – 2y – 4) (x2 – 2y + 2) (a)

11. The factors of x3 – 7x + 6 are

(a) x(x – 6) (x – 1) (b) (x2 – 6) (x – 1)

(c) (x + 1) (x + 2) (x – 3) (d) (x – 1) (x + 3) (x – 2)

Sol. x3 – 7x + 6 = x3 – 1 – 7x + 7

= (x – 1) (x2 + x + 1) – 7(x – 1)

= (x – 1) (x2 + x + 1 – 7) = (x – 1) (x2 + x – 6)

= (x – 1) [x2 + 3x – 2x – 6]

= (x – 1) [x(x + 3) – 2(x + 3)]

= (x – 1) (x + 3) (x – 2) (d)

12. The expression x4 + 4 can be factorized as

(a) (x2 + 2x + 2) (x2 – 2x + 2) (b) (x2 + 2x + 2) (x2 + 2x – 2)

(c) (x2 – 2x – 2) (x2 – 2x + 2) (d) (x2 + 2) (x2 – 2)

Sol. x4 + 4 = x4 + 4 + 4x2 – 4x2 (Adding and subtracting 4x2)

= (x2)2 + (2)2 + 2 × x2 × 2 – (2x)2

= (x2 + 2)2 – (2x)2

= (x2 + 2 + 2x) (x2 + 2 – 2x) {_ a2 – b2 = (a + b) (a – b)}

= (x2 + 2x + 2) (x2 – 2x + 2) (a)

13. If 3x = a + b + c, then the value of (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c) is

(a) a + b + c (b) (a – b) (b – c) (c – a)


Sol. 3x = a + b + c

3x – a – b – c = 0

Now, (x – a)3 + (x – b)3 + (x – c)3 – 3(x – a) (x – b) (x – c)

= {(x – a) + (x – b) + (x – c)} {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b)

– (x – b) (x – c) – (x – c) (x – a)}

= (x – a + x – b + x – c) {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b)

– (x – b) (x – c) – (x – c) (x – a)}

= (3x – a – b – c) {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}



But 3x – a – b – c = 0, then

= 0 × {(x – a)2 + (x – b)2 + (x – c)2 – (x – a) (x – b) – (x – b) (x – c) – (x – c) (x – a)}

= 0 (c)

14. If (x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2, then k =

(a) 1 (b) 2 (c) 4 (d) 8

Sol. (x + y)3 – (x – y)3 – 6y(x2 – y2) = ky2

LHS = (x + y)3 – (x – y)3 – 3 × (x + y) (x – y) [x + y – x + y]

= (x + y – x + y)3 {_ a3 – b3 – 3ab (a – b) = a3 – b3}

= (2y)3 = 8y3

Comparing with ky3, k = 8 (d)

15. If x3 – 3x2 + 3x – 7 = (x + 1) (ax2 + bx + c), then a + b + c =

(a) 4 (b) 12 (c) –10 (d) 3

Sol. x3 – 3x2 + 3x + 7 = (x + 1) (ax2 + bx + c)

= ax3 + bx2 + cx + ax2 + bx + c

x3 – 3x2 + 3x – 7 = ax3 + (b + a)x2 + (c + b)x + c

Comparing the coefficient,

a = 1

b + a = –3 b + 1 = –3 b = –3 – 1 = –4

c + b = 3 c – 4 = 3 c = 3 + 4 = 7

a + b + c = 1 – 4 + 7 = 8 – 4 = 4 (a)


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· Exercise 1. 3 -3 Exercise 1. 3 -3 er r s er e e si s 3 -i e ice e si s - 1 1 16. stru t 1- 1 Exercise 1.1 1-1 Exercise 1. 1-1 Exercise 1.3 1 -1 1. $er"s rmu -