# ML Aggarwal Expansions Exe-3.2 Class 9 ICSE Maths Solutions

ML Aggarwal Expansions Exe-3.2 Class 9 ICSE Maths APC Understanding Solutions . Solutions of  Exercise-3.2. This post is the Solutions of  ML Aggarwal  Chapter 3  – Expansions  for ICSE Maths Class-9.  APC Understanding ML Aggarwal Solutions (APC) Avichal Publication Solutions of Chapter-3.2 Expansions for ICSE Board Class-9. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Expansions Exe-3.2 Class 9 ICSE Maths Solutions

 Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 9 th Chapter-3 Expansions (Exercise 3.2) Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-3.2 Questions Academic Session 2021-2022

### Exe-3.2 Solutions of ML Aggarwal for ICSE Class-9 Chapter-3, Expansions

Note:- Before viewing Solutions of Chapter -3 Expansions Class-9 of ML Aggarwal Solutions .  Read the Chapter Carefully then solve all example given in  Exercise-3.2 . The Chapter-3 Expansions Class-9 is Main Chapter in Class 9 Mathematics.

### Expansions Exe-3.2

ML Aggarwal Class 9 ICSE Maths Solutions

Page 84

#### Question 1. If x – y = 8 and xy = 5, find x2 + y2.

We know that

(x – y)2 = x2 + y2 – 2xy

It can be written as

x2 + y2 = (x – y)2 + 2xy

It is given that

x – y = 8 and xy = 5

Substituting the values

x2 + y2 = 82 + 2 × 5

So we get

= 64 + 10

= 74

#### Question 2. If x + y = 10 and xy = 21, find 2 (x2 + y2).

We know that

(x + y)2 = x2 + y2 + 2xy

It can be written as

x2 + y2 = (x + y)2 – 2xy

It is given that

(x + y) = 10 and xy = 21

Substituting the values

x2 + y2 = 102 – 2 × 21

By further calculation

= 100 – 42

= 58

Here

2 (x2 + y2) = 2 × 58 = 116

#### Question 3. If 2a + 3b = 7 and ab = 2, find 4a2 + 9b2.

We know that

(2a + 3b)2 = 4a2 + 9b2 + 12ab

It can be written as

4a2 + 9b2 = (2a + 3b)2 – 12ab

It is given that

2a + 3b = 7

ab = 2

Substituting the values

4a2 + 9b2 = 72 – 12 × 2

By further calculation

= 49 – 24

= 25

#### Question 4. If 3x – 4y = 16 and xy = 4, find the value of 9x2 + 16y2.

We know that

(3x – 4y)2 = 9x2 + 16y2 – 24xy

It can be written as

9x2 + 16y2 = (3x – 4y)2 + 24xy

It is given that

3x – 4y = 16 and xy = 4

Substituting the values

9x2 + 16y2 = 162 + 24 × 4

By further calculation

= 256 + 96

= 352

#### Question 5. If x + y = 8 and x – y = 2, find the value of 2x2 + 2y2.

We know that

2 (x2 + y2) = (x + y)2 + (x – y)2

It is given that

x + y = 8 and x – y = 2

Substituting the values

2x2 + 2y2 = 82 + 22

By further calculation

= 64 + 4

= 68

#### Question 6. If a2 + b2 = 13 and ab = 6, find

(i) a + b

(ii) a – b

##### (i) We know that

(a + b)2 = a2 + b2 + 2ab

Substituting the values

= 13 + 2 × 6

So we get

= 13 + 12

= 25

Here

a + b = ± √25 = ± 5

##### (ii) We know that

(a – b)2 = a2 + b2 – 2ab

Substituting the values

= 13 – 2 × 6

So we get

= 13 – 12

= 1

Here

a – b = ± √1 = ± 1

#### Question 7. If a + b = 4 and ab = -12, find

(i) a – b

(ii) a2 – b2.

##### (i) We know that

(a – b)2 = a2 + b2 – 2ab

It can be written as

(a – b)2 = a2 + b2 + 2ab – 4ab

(a – b)2 = (a + b)2 – 4ab

It is given that

a + b = 4 and ab = – 12

Substituting the values

(a – b)2 = 42 – 4 (-12)

By further calculation

(a – b)2 = 16 + 48 = 64

So we get

(a – b) = ± √64 = ± 8

##### (ii) We know that

a2 – b2 = (a + b) (a – b)

Substituting the values

a2 – b2 = 4 × ±8

a2 – b2 = ± 32

#### Question 8. If p – q = 9 and pq = 36, evaluate

##### (i) We know that

(p + q)2 = p2 + q2 + 2pq

It can be written as

(p + q)2 = p2 + q2 – 2pq + 4pq

(p + q)2 = (p – q)2 + 4pq

It is given that

p – q = 9 and pq = 36

Substituting the values

(p + q)2 = 92 + 4 × 36

By further calculation

(p + q)2 = 81 + 144 = 225

So we get

p + q = ± √225 = ± 15

##### (ii) We know that

p2 – q2 = (p – q) (p + q)

Substituting the values

p2 – q2 = 9 ×±15

p2 – q2 = ± 135

#### Question 9. If x + y = 6 and x – y = 4, find

(i) x2 + y2

(ii) xy

We know that

(x + y)2 – (x – y)2 = 4xy

Substituting the values

62 – 42 = 4xy

By further calculation

36 – 16 = 4xy

20 = 4xy

4xy = 20

So we get

xy = 20/4 = 5

##### (i) x2 + y2 = (x + y)2 – 2xy

Substituting the values

= 62 – 2 × 5

By further calculation

= 36 – 10

= 26

#### Question 10. If x – 3 = 1/x, find the value of x2 + 1/x2.

It is given that

x – 3 = 1/x

We can write it as

x – 1/x = 3

Here

(x – 1/x)2 = x2 + 1/x2 – 2

So we get

x2 + 1/x2 = (x – 1/x)2 + 2

Substituting the values

x2 + 1/x2 = 32 + 2

By further calculation

= 9 + 2

= 11

#### Question 11. If x + y = 8 and xy = 3 ¾, find the values of

(i) x – y

(ii) 3 (x2 + y2)

(iii) 5 (x2 + y2) + 4 (x – y).

##### (i) We know that

(x – y)2 = x2 + y2 – 2xy

It can be written as

(x – y)2 = x2 + y2 + 2xy – 4xy

(x – y)2 = (x + y)2 – 4xy

It is given that

x + y = 8 and xy = 3 ¾ = 15/4

Substituting the values

(x – y)2 = 82 – 4 × 15/4

So we get

(x – y)2 = 64 – 15 = 49

x – y = ± √49 = ± 7

##### (ii) We know that

(x + y)2 = x2 + y2 + 2xy

We can write it as

x2 + y2 = (x + y)2 – 2xy

It is given that

x + y = 8 and xy = 3 ¾ = 15/4

Substituting the values

x2 + y2 = 82 – 2 × 15/4

So we get

x2 + y2 = 64 – 15/2

Taking LCM

x2 + y2 = (128 – 15)/ 2 = 113/2

We get

3 (x2 + y2) = 3 × 113/2 = 339/2 = 169 ½

##### (iii) We know that

5 (x2 + y2) + 4 (x – y) = 5 × 113/2 + 4 × ± 7

By further calculation

= 565/2 ± 28

We can write it as

= 565/2 + 28 or 565/2 – 28

= 621/2 or 509/2

It can be written as

= 310 ½ or 254 ½

#### Question 12. If x2 + y2 = 34 and xy = 10 ½, find the value of 2 (x + y)2 + (x – y)2.

It is given that

x2 + y2 = 34 and xy = 10 ½ = 21/2

We know that

(x + y)2 = x2 + y2 + 2xy

Substituting the values

(x + y)2 = 34 + 2 (21/2)

So we get

(x + y)2 = 55 ….. (1)

We know that

(x – y)2 = x2 + y2 – 2xy

Substituting the values

(x – y)2 = 34 – 2 (21/2)

So we get

(x – y)2 = 34 – 21 = 13 ….. (2)

Using both the equations

2 (x + y)2 + (x – y)2 = 2 × 55 + 13 = 123

#### Question 13. If a – b = 3 and ab = 4, find a3 – b3.

We know that

a3 – b3 = (a – b)3 + 3ab (a + b)

Substituting the values

a3 – b3 = 33 + 3 × 4 × 3

By further calculation

a3 – b3 = 27 + 36 = 63

#### Question 14. If 2a – 3b = 3 and ab = 2, find the value of 8a3 – 27b3.

We know that

8a3 – 27b3 = (2a)3 – (3b)3

According to the formula

= (2a – 3b)3 + 3 × 2a × 3b (2a – 3b)

By further simplification

= (2a – 3b)3 + 18ab (2a – 3b)

Substituting the values

= 33 + 18 × 2 × 3

By further calculation

= 27 + 108

= 135

#### Question 15. If x + 1/x = 4, find the values of

(i) x2 + 1/x2

(ii) x4 + 1/x4

(iii) x3 + 1/x3

(iv) x – 1/x.

##### (i) We know that

(x + 1/x)2 = x2 + 1/x2 + 2

It can be written as

x2 + 1/x2 = (x + 1/x)2 – 2

Substituting the values

= 42 – 2

= 16 – 2

= 14

##### (ii) We know that

(x2 + 1/x2)2 = x4 + 1/x4 + 2

It can be written as

x4 + 1/x4 = (x2 + 1/x2)2 – 2

Substituting the values

= 142 – 2

= 196 – 2

= 194

##### (iii) We know that

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

It can be written as

(x + 1/x)3 – 3(x + 1/x) = 43 – 3 × 4

By further calculation

= 64 – 12

= 52

##### (iv) We know that

(x – 1/x)2 = x2 + 1/x2 – 2

Substituting the values

= 14 – 2

= 12

So we get

x – 1/x = ± 2√3

#### Question 16. If x – 1/x = 5, find the value of x4 + 1/x4.

We know that

(x – 1/x)2 = x2 + 1/x2 – 2

It can be written as

x2 + 1/x2 = (x – 1/x)2 + 2

Substituting the values

x2 + 1/x2 = 52 + 2 = 27

Here

x4 + 1/x4 = (x2 + 1/x2)2 – 2

Substituting the values

x4 + 1/x4 = 272 – 2

So we get

= 729 – 2

= 727

#### Question 17. If x – 1/x = √5, find the values of

(i) x2 + 1/x2

(ii) x + 1/x

(iii) x3 + 1/x3

##### (i) x2 + 1/x2 = (x – 1/x)2 + 2

Substituting the values

= (√5)2 + 2

= 5 + 2

= 7

##### (ii) (x + 1/x)2 = x2 + 1/x2 + 2

Substituting the values

= 7 + 2

= 9

Here

(x + 1/x)2 = 9

So we get

(x + 1/x) = ± √9 = ± 3

##### (iii) x3 + 1/x3 = (x + 1/x)3 – 3x (1/x) (x + 1/x)

Substituting the values

= (± 3)3 – 3 (± 3)

By further calculation

= (± 27) – (± 9)

= ± 18

#### Question 18. If x + 1/x = 6, find

(i) x – 1/x

(ii) x2 – 1/x2.

##### (i) We know that

(x – 1/x)2 = x2 + 1/x2 – 2

It can be written as

(x – 1/x)2 = x2 + 1/x2 + 2 – 4

(x – 1/x)2 = (x + 1/x)2 – 4

Substituting the values

(x – 1/x)2 = 62 – 4 = 32

So we get

x – 1/x = ± √32 = ± 4√2

##### (ii) We know that

x2 – 1/x2 = (x – 1/x) (x + 1/x)

Substituting the values

x2 – 1/x2 = (± 4√2) (6) = ± 24 √2

#### Question 19. If x + 1/x = 2, prove that x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4.

We know that

x2 + 1/x2 = (x + 1/x)² – 2

Substituting the values

x2 + 1/x2 = 22 – 2

So we get

x2 + 1/x2 = 4 – 2 = 2 …. (1)

x3 + 1/x3 = (x + 1/x)3 – 3 (x + 1/x)

Substituting the values

x3 + 1/x3 = 23 – 3 × 2

So we get

x3 + 1/x3 = 8 – 6 = 2 …… (2)

x4 + 1/x4 = (x2 + 1/x2)2 – 2

Substituting the values

x4 + 1/x4 = 22 – 2

So we get

x4 + 1/x4 = 4 – 2 = 2 …. (3)

From equation (1), (2) and (3)

x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4

Hence, it is proved.

#### Question 20. If x – 2/x = 3, find the value of x3 – 8/x3.

We know that

(x – 2/x)3 = x3 – 8/x3 – 3 (x) (2/x) (x – 2/x)

By further simplification

(x – 2/x)3 = x3 – 8/x3 – 6 (x – 2/x)

It can be written as

x3 – 8/x3 = (x – 2/x)3 + 6 (x – 2/x)

Substituting the values

x3 – 8/x3 = 33 + 6 × 3

By further calculation

x3 – 8/x3 = 27 + 18 = 45

#### Question 21. If a + 2b = 5, prove that a3 + 8b3 + 30ab = 125.

We know that

(a + 2b)3 = a3 + 8b3 + 3 (a) (2b) (a + 2b)

Substituting the values

53 = a3 + 8b3 + 6ab (5)

By further calculation

125 = a3 + 8b3 + 30ab

Therefore, a3 + 8b3 + 30ab = 125.

ML Aggarwal Class 9 ICSE Maths Solutions

Page 85

#### Question 22. If a + 1/a = p, prove that a3 + 1/a3 = p (p2 – 3).

We know that

a3 + 1/a3 = (a + 1/a)3 – 3a (1/a) (a + 1/a)

Substituting the values

a3 + 1/a3 = p3 – 3p

Taking p as common

a3 + 1/a3 = p (p2 – 3)

Therefore, it is proved.

#### Question 23. If x2 + 1/x2 = 27, find the value of x – 1/x.

We know that

(x – 1/x)2 = x2 + 1/x2 – 2

Substituting the values

(x – 1/x)2 = 27 – 2 = 25

So we get

x – 1/x = ± √25 = ± 5

#### Question 24. If x2 + 1/x2 = 27, find the value of 3x3 + 5x – 3/x3 – 5/x.

We know that

(x – 1/x)2 = x2 + 1/x2 – 2

Substituting the values

(x – 1/x)2 = 27 – 2 = 25

So we get

x – 1/x = ± √25 = ± 5

Here

3x3 + 5x – 3/x3 – 5/x = 3 (x3 – 1/x3) + 5 (x – 1/x)

It can be written as

= 3 [(x – 1/x)3 + 3 (x – 1/x)] + 5 (x – 1/x)

Substituting the values

= 3 [(± 5)3 + 3 (± 5)] + 5 (± 5)

By further calculation

= 3 [(± 125) + (± 15)] + (± 25)

So we get

= (± 375) + (± 45) + (± 25)

= ± 445

#### Question 25. If x2 + 1/25x2 = 8 3/5, find x + 1/5x.

We know that

(x + 1/5x)2 = x2 + 1/25x2 + 2x (1/5x)

It can be written as

(x + 1/5x)2 = x2 + 1/25x2 + 2/5

Substituting the values

(x + 1/5x)2 = 8 3/5 + 2/5

(x + 1/5x)2 = 43/5 + 2/5

So we get

(x + 1/5x)2 = 45/5 = 9

Here

x + 1/5x = ± √9 = ± 3

#### Question 26. If x2 + 1/4x2 = 8, find x3 + 1/8x3.

We know that

(x + 1/2x)2 = x2 + (1/2x)2 + 2x (1/2x)

It can be written as

(x + 1/2x)2 = x2 + 1/4x2 + 1

Substituting the values

(x + 1/2x)2 = 8 + 1 = 9

So we get

x + 1/2x = ± √9 = ± 3

Here

x3 + 1/8x3 = x3 + (1/2x)3

We know that

x3 + 1/8x3 = (x + 1/2x)3 – 3x (1/2x) (x + 1/2x)

Substituting the values

x3 + 1/8x3 = (± 3)3 – 3/2 (± 3)

By further calculation

x3 + 1/8x3 = ± (27 – 9/2)

Taking LCM

x3 + 1/8x3 = ± (54 – 9)/ 2

x3 + 1/8x3 = ± 45/2 = ± 22 ½

Therefore, x3 + 1/8x3 = ± 22 ½.

#### Question 27. If a2 – 3a + 1 = 0, find

(i) a2 + 1/a2

(ii) a3 + 1/a3.

It is given that

a2 – 3a + 1 = 0

By dividing each term by a

a + 1/a = 3

##### (i) We know that

(a + 1/a)2 = a2 + 1/a2 + 2

It can be written as

a2 + 1/a2 = (a + 1/a)2 – 2

Substituting the values

= 32 – 2

= 9 – 2

= 7

##### (ii) We know that

(a + 1/a)3 = a3 + 1/a3 + 3 (a + 1/a)

It can be written as

a3 + 1/a3 = (a + 1/a)3 – 3 (a + 1/a)

Substituting the values

= 33 – 3 (3)

= 27 – 9

= 18

#### Question 28. If a = 1/ (a – 5), find

(i) a – 1/a

(ii) a + 1/a

(iii) a2 – 1/a2.

It is given that

a = 1/ (a – 5)

We can write it as

a2 – 5a – 1 = 0

Now divide each term by a

a – 5 – 1/a = 0

So we get

a – 1/a = 5

##### (ii) We know that

(a + 1/a)2 = (a – 1/a)2 + 4

Substituting the values

(a + 1/a)2 = 52 + 4

So we get

(a + 1/a)2 = 25 + 4 = 29

a + 1/a = ± √29

##### (iii) We know that

a2 – 1/a2 = (a + 1/a) (a – 1/a)

Substituting the values

a2 – 1/a2 = ± √29 × 5

a2 – 1/a2 = ± 5√29

#### Question 29. If (x + 1/x)2 = 3, find x3 + 1/x3.

It is given that

(x + 1/x)2 = 3

(x + 1/x) = ± √3

We know that

x3 + 1/x3 = (x + 1/x)3 – 3 (x + 1/x)

Substituting the values

x3 + 1/x3 = (± √3)3 – 3 (± √3)

By further calculation

x3 + 1/x3 = (± 3√3) – (± 3√3) = 0

#### Question 30. If x = 5 – 2√6, find the value of √x + 1/√x.

It is given that

x = 5 – 2√6

We can write it as

= 5 + 2√6

Here

x + 1/x = 5 – 2√6 + 5 + 2√6 = 10

So we get

(√x + 1/√x)2 = x + 1/x + 2

Substituting the values

= 10 + 2

= 12

#### Question 31. If a + b + c = 12 and ab + bc + ca = 22, find a2 + b2 + c2.

We know that

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

We can write it as

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

Substituting the values

a2 + b2 + c2 = 122 – 2 (22)

By further calculation

a2 + b2 + c2 = 144 – 44 = 100

#### Question 32. If a + b + c = 12 and a2 + b2 + c2 = 100, find ab + bc + ca.

We know that

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

It can be written as

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

Taking out 2 as common

2 (ab + bc + ca) = 122 – 100 = 144 – 100 = 44

By further calculation

ab + bc + ca = 44/2 = 22

#### Question 33. If a2 + b2 + c2 = 125 and ab + bc + ca = 50, find a + b + c.

We know that

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

Substituting the values

(a + b + c)2 = 125 + 2 (50)

By further calculation

(a + b + c)2 = 125 + 100 = 225

So we get

a + b + c = ± √225 = ± 15

#### Question 34. If a + b – c = 5 and a2 + b2 + c2 = 29, find the value of ab – bc – ca.

It is given that

a + b – c = 5

By squaring on both sides

(a + b – c)2 = 52

Expanding using formula

a2 + b2 + c2 + 2ab – 2bc – 2ca = 25

Substituting the values and taking 2 as common

29 + 2 (ab – bc – ca) = 25

By further calculation

2 (ab – bc – ca) = 25 – 29 = – 4

So we get

ab – bc – ca = – 4/2 = – 2

Therefore, ab – bc – ca = – 2.

#### Question 35. If a – b = 7 and a2 + b2 = 85, then find the value of a3 – b3.

We know that

(a – b)2 = a2 + b2 – 2ab

Substituting the values

72 = 85 – 2ab

By further calculation

49 = 85 – 2ab

So we get

2ab = 85 – 49 = 36

Dividing by 2

ab = 36/2 = 18

Here

a3 – b3 = (a – b) (a2 + b2 + ab)

Substituting the values

a3 – b3 = 7 (85 + 18)

By further calculation

a3 – b3 = 7 × 103

So we get

a3 – b3 = 721

#### Question 36. If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.

It is given that

x = y – 3 and x2 + y2 = 29

It can be written as

x – y = – 3

By squaring on both sides

(x – y)2 = (-3)2

Expanding using formula

x2 + y2 – 2xy = 9

Substituting the values

29 – 2xy = 9

By further calculation

-2xy = 9 – 29 = – 20

Dividing by 2

xy = – 20/-2 = 10

So we get

xy = 10

#### Question 37. If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.

Consider x and y as the two numbers

x + y = 8 and xy = 15

By cubing on both sides

(x + y)3 = 83

Expanding using formula

x3 + y3 + 3xy (x + y) = 512

Substituting the values

x3 + y3 + 3 × 15 × 8 = 512

By further calculation

x3 + y3 + 360 = 512

So we get

x3 + y3 = 512 – 360 = 152

—  : End of ML Aggarwal Expansions Exe-3.2 Class 9 ICSE Maths Solutions :–

Thanks

### 3 thoughts on “ML Aggarwal Expansions Exe-3.2 Class 9 ICSE Maths Solutions”

1. there is a error in qna 11 (i) instead of Substituting the values

(x – y)2 = 82 – 4 × 15/4

So we get

(x – y)2 = ’65’ – 15 = 49

x – y = ± √49 = ± 7

it should be

Substituting the values

(x – y)2 = 82 – 4 × 15/4

So we get

(x – y)2 = ’64’ – 15 = 49

x – y = ± √49 = ± 7