# Frank Solutions for Chapter 4 Expansions Class 9 Mathematics ICSE

### Exercise 4.1

1. Expand the following:

(i) (a + 4) (a + 7)

(ii) (m + 8) (m – 7)

(iii) (x – 5) (x – 4)

(iv) (3x + 4) (2x – 1)

(v) (2x – 5) (2x + 5) (2x – 3)

(i) (a + 4) (a + 7)

= a+ 4a + 7a + 28

We get,

= a2 + 11a + 28

(ii) (m + 8) (m – 7)

= m2 + 8m – 7m – 56

We get,

= m2 + m – 56

(iii) (x – 5) (x – 4)

= x2 – 5x – 4x + 20

We get,

= x2 – 9x + 20

(iv) (3x + 4) (2x – 1)

= 6x2 – 3x + 8x – 4

We get,

= 6x2 + 5x – 4

(v) (2x – 5) (2x + 5) (2x – 3)

= (4x– 25) (2x – 3)

We get,

= 8x3 – 12x2 – 50x + 75

2. Expand the following:

(i) (a + 3b)2

(ii) (2p – 3q)2

(iii) (2a + 1/2a)2

(iv) (x – 3y – 2z)2

(i) (a + 3b)2

(a + 3b)2 = a2 + 2(a) (3b) + (3b)2 [Using (x + y)2 = x2 + 2xy + y2]

We get,

= a2 + 6ab + 9b2

(ii) (2p – 3q)2

= (2p)2 – 2 (2p) (3q) + (3q)2

We get,

= 4p2 – 12pq + 9q2

(iii) (2a + 1/2a)2

= (2a)2 + 2 (2a) (1/2a) + (1/2a)2

We get,

= {4a2 + 2 + (1/4a2)}

(iv) (x – 3y – 2z)2

= x2 + (3y)2 + (2z)2 + 2 (x) (-3y) + 2 (-3y) (-2z) + 2 (x) (-2z)

[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ac]

We get,

= x2 + 9y2 + 4z2 – 6xy + 12yz – 4xz

3. Find the squares of the following:

(a) 9m – 2n

(b) 3p – 4q

(c) (7x/9y) – (9y/7x)

(d) (2a + 3b – 4c)

(a) 9m – 2n

(9m – 2n)2 = (9m)2 + 2 (9m) (-2n) + (-2n)2

[Using (x + y)2 = x2 + 2xy + y2]

We get,

= 81m2 – 36mn + 4n2

(b) 3p – 4q

(3p – 4q)2 = (3p)2 – 2 (3p) (4q) + (4q)2

On simplification, we get,

= 9p2 – 12pq + 16q2

(c) {(7x/9y) – (9y/7x)}2 = (7x/9y)2 + 2 (7x/9y) (9y/7x) + (9y/7x)2

On simplification, we get,

= (49x2/81y2) + 2 + (81y2/49x2)

(d) (2a + 3b – 4c)

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

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

On simplification, we get,

= 4a2 + 9b2 + 16c2 + 12ab -24bc – 16ac

4. Simplify by using formula:

(i) (5x – 9) (5x + 9)

(ii) (2x + 3y) (2x – 3y)

(iii) (a + b – c) (a – b + c)

(iv) (x + y – 3) (x + y + 3)

(v) (1 + a) (1 – a) (1 + a2)

(vi) {a + (2/a) – 1}{a – (2/a) – 1}

(i) (5x – 9) (5x + 9)

[Using identity: (a + b) (a – b) = a2 – b2]

= (5x)2 – (9)2

We get,

= 25x2 – 81

(ii) (2x + 3y) (2x – 3y)

[Using identity: (a + b) (a – b) = a2 – b2]

= (2x)2 – (3y)2

We get,

= 4x2 – 9y2

(iii) (a + b – c) (a – b + c)

On further calculation, we get,

= (a + b – c) {a – (b – c)}

[Using identity: (a + b) (a – b) = a2 – b2]

= (a)2 – (b – c)2

= a2 – (b2 + c2 – 2bc)

= a2 – b2 – c2 + 2bc

(iv) (x + y – 3) (x + y + 3)

[Using identity: (a + b) (a – b) = a2 – b2]

= (x + y)2 – (3)2

= x2 + y2 + 2xy – 9

(v) (1 + a) (1 – a) (1 + a2)

[Using identity: (a + b) (a – b) = a2 – b2]

= {(1)2 – (a)2} (1 + a2)

= (1 – a2) (1 + a2)

= (1)2 – (a2)2

We get,

= 1 – a4

(vi) {a + (2 / a) – 1} {a – (2 / a) – 1}

[Using identity: (a + b) (a – b) = a2 – b2]

= (a – 1)2 – (2 / a)2

We get,

= a2 + 1 – 2a – (4 / a2)

5. Evaluate the following without multiplying:

(i) (95)2

(ii) (103)2

(iii) (999)2

(iv) (1005)2

(i) (95)2

Using (x + y)2 = x2 + 2xy + y2

We get,

(95)2 = (100 – 5)2

= (100)2 – 2 (100) (5) + (5)2

= 10000 – 1000 + 25

We get,

= 9025

(ii) (103)2

Using (x + y)2 = x2 + 2xy + y2

We get,

(103)2 = (100 + 3)2

= (100)2 + 2 (100) (3) + (3)2

= 10000 + 600 + 9

We get,

= 10609

(iii) (999)2

(999)2 = (1000 – 1)2

= (1000)2 – 2 (1000) (1) + (1)2

= 1000000 – 2000 + 1

We get,

= 998001

(iv) (1005)2

(1005)2 = (1000 + 5)2

= (1000)2 + 2 (1000) (5) + (5)2

= 1000000 + 10000 + 25

We get,

= 1010025

6. Evaluate, using (a + b) (a – b) = a2– b2

(i) 399 × 401

(ii) 999 × 1001

(iii) 409 × 5.1

(iv) 15.9 × 16.1

(i) 399 × 401

399 × 401 = (400 – 1) (400 + 1)

= (400)2 – (1)2

On further calculation, we get,

= 160000 – 1

We get,

= 159999

(ii) 999× 1001

999 × 1001 = (1000 – 1) (1000 + 1)

= (1000)2 – (1)2

On further calculation, we get,

= 1000000 – 1

We get,

= 999999

(iii) 4.9 × 5.1

4.9 × 5.1 = (5 – 0.1) (5 + 0.1)

= (5)2 – (0.1)2

= 25 – 0.01

We get,

= 24.99

(iv) 15.9 × 16.1

15.9 × 16.1 = (16 – 0.1) (16 + 0.1)

= (16)2 – (0.1)2

= 256 – 0.01

We get,

= 255.99

7. If a – b = 10 and ab = 11; find a + b

Given

a – b = 10 and ab = 11

We know that,

(a – b)2 = a2 – 2ab + b2

⇒ (10)2 = a2 – 2 (11) + b2

⇒ 100 = a2 + b2 – 22

On further calculation, we get,

a2 + b2 = 100 + 22

⇒ a2 + b2 = 122

[Using (a + b)2 = a2 + b2 + 2ab]

We get,

(a + b)2 = 122 + 2 (11)

⇒ (a + b)2 = 122 + 22

⇒ (a + b)2 = 144

⇒ (a + b) = √144

We get,

(a + b) = ± 12

8. If x + y = 9, xy = 20; find:

(i) x – y

(ii) x2 – y2

(i) Given

x + y = 9, xy = 20

We know that,

(a + b)2 = a2 + 2ab + b2

(x + y)2 = x2 + y2 + 2xy

⇒ x2 + y2 + 2xy = 81

⇒ x2 + y2 = 81 – 2(xy)

⇒ x2 + y2 = 81 – 2 (20)

⇒ x2 + y2 = 81 – 40

⇒ x2 + y2 = 41

We know that,

(a – b)2 = a2 – 2ab + b2

(x – y)2 = x2 – 2xy + y2

⇒ (x – y)2 = x2 + y2 – 2xy

⇒ (x – y)2 = 41 – 2 (20)

⇒ (x – y)2 = 41 – 40

⇒ (x – y)= 1

⇒ x – y = ± 1

(ii) We know that,

(x – y) (x + y) = x2 – y2

⇒ x2 – y2 = (± 1) (9)

We get,

x2 – y2 = ± 9

9. Find the cube of:

(i) 2a – 5b

(ii) 4x + 7y

(iii) 3a + (1/3a)

(iv) 4p – (1/p)

(v) (2m/3n) + (3n/2m)

(vi) a – (1/a) + b

(i) 2a – 5b

Using, (a – b)3 = a3 – b3 – 3ab (a – b)

(2a – 5b)3 = (2a)3 – (5b)3 – 3 (2a) (5b) (2a – 5b)

On further calculation, we get,

= 8a3 – 125b3 – 30ab (2a – 5b)

= 8a3 – 125b3 – 60a2b + 150ab2

(ii) 4x + 7y

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

(4x + 7y)3 = (4x)3 + (7y)3 + 3 (4x) (7y) (4x + 7y)

On further calculation, we get,

= 64x3 + 343y3 + 84xy (4x + 7y)

= 64x3 + 343y3 + 336x2y + 588xy2

(iii) {3a + (1/3a)}3

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

{3a + (1/3a)}3 = (3a)3 + (1/3a)3 + 3 (3a) (1/3a) {3a + (1/3a)}

We get,

= 27a3 + (1/27a3) + 9a + (1/a)

(iv) {4p – (1/p)}3

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

{4p – (1/p)}3 = (4p)3 – (1/p)3 – 3 (4p) (1/p) {4p – (1/p)}

We get,

= 64p3 – (1/p3) – 48p + (12/p)

(v) {(2m/3n) + (3n/2m)}3

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

= (2m/3n)3 + (3n/2m)3 + {3 (2m/3n) (3n/2m)}{(2m/3n) + (3n/2m)}

We get,

= (8m3/27n3) + (27n3/8m3) + (2m/n) + (9n/2m)

(vi) {a – (1/a) + b}3

Using (a + b + c)3 = a3 + b3 + c3 + 3a2b + 3a2c + 3b2a + 3c2a + 6abc

{a – (1/a) + b}3 = a3 + (-1/a)3 + b3 + 3a2 (-1/a) + 3a2b + 3 (-1 /a)2b + 3(-1/a)2a + 3b2a + 3b2(-1/a)+ 6a(-1/a)b

We get,

= a3 – (1/a3) + b3 – 3a + 3a2b + 3b/a2 + 3/a + 3b2a – 3b2/a – 6b

10. If {5x + (1/5x)} = 7; find the value of 125x3+ (1/125x3)

Given

{5x + (1/5x)} = 7

Using {a + (1/a)}3 = a3 + (1/a3) + 3 {a + (1/a)}

We get,

{5x + (1 / 5x)}3 = (5x)3 + (1/5x)3 + 3 {5x + (1/5x)}

⇒ 343 = 125x3 + (1/125x3) + 3 (7)

We get,

125x3 + (1/125x3) = 343 – 21

⇒ 125x3 + (1/125x3) = 322

11. If {3x – (1 / 3x)} = 9; find the value of 27x3– (1 / 27x3)

Given

{3x – (1/3x)} = 9

Using {a – (1/a)}3 = a3 – (1/a3) – 3 {a – (1/a)}

We get,

{3x – (1/3x)}3 = (3x)3 – (1/3x)3 – 3 {3x – (1/3x)}

⇒ 729 = 27x3 – (1 /27x3) – 3(9)

On calculating further, we get,

27x3 – (1/27x3) = 729 + 27

⇒ 27x3 – (1/27x3) = 756

12. If (x + 1/x) = 5, find the value of {x2+ (1/x2), x3+ (1/x3) and x4 + (1/x4)}

Given

(x + 1/x) = 5 …(1)

On squaring both sides,

We get,

(x + 1/x)2 = (5)2

⇒ x2 + (1/x2) + 2 = 25

⇒ x2 + (1/x2) = 25 – 2

⇒ x2 + (1/x2) = 23 …..(2)

Now,

Cubing both sides of equation (1),

We get,

{x + (1/x)}3 = (5)3

⇒ x3 + (1/x3) + 3 {x + (1/x)} = 125

⇒ x3 + (1/x3) + 3(5) = 125

On calculating further, we get,

x3 + (1/x3) = 125 – 15

⇒ x3 + (1/x3) = 110

Squaring on both sides of equation (2), we get,

{x2 + (1/x2)}2 = (23)2

⇒ x4 + (1/x4) + 2 = 529

⇒ x4 + (1/x4) = 529 – 2

⇒ x4 + (1/x4) = 527

13. If {a – (1/a)} = 7, find {a2+ (1/a2)}, {a2– (1/a2) and {a3 – (1/a3)}

Given

{a – (1/a) = 7 ….(1)

Squaring on both sides, we get,

{a – (1/a)}2 = (7)2

⇒ a2 + (1/a2) – 2 = 49

⇒ a2 + (1/a2) = 49 + 2

⇒ a2 + (1/a2) = 51

Now,

{a + (1/a)}2 = a2 + (1/a2) + 2

Substitute the value of a2 + (1/a)2, we get,

= 51 + 2

= 53

a + (1 / a) = ± √53

Now,

a2 – (1/a2) = {a + (1/a)}{a – (1/a)}

= (± √53) (7)

We get,

= (± 7√53)

Cubing on both sides of equation (1), we get,

{a – (1/a)}3 = (7)3

⇒ a3 – (1/a3) – 3 {a – (1/a)} = 343

⇒ a3 – (1/a3) – 3 (7) = 343

⇒ a3 – (1/a3) = 343 + 21

⇒ a3 – (1/a3) = 364

14. If {a2+ (1 / a2)} = 14; find the value of

(i) {a + (1/a)}

(ii) {a3 + (1/a3)}

(i) Using (a + b)2 = a2 + 2ab + b2

{a + (1/a)}2 = a2 + 2a (1/a) + (1/a)2

⇒ {a + (1/a)}2 = a2 + 2 + (1/a2)

⇒ {a + (1/a)}2 = a2 + (1/a2) + 2

Substitute the value of {a2 + (1 / a2)}, we get,

{a + (1/a)}2 = 14 + 2

⇒ {a + (1/a)}2 = 16

⇒ {a + (1/a)} = ± 4

Therefore, the value of {a + (1/a)} = ± 4

(ii) {a+ (1/a3)} = {a + (1/a)} {(a2 + (1/a2) – 1}

[Using a3 + b3 = (a + b) (a2 + b2 – ab)]

We get,

{a3 + (1/a3)} = (± 4) (14 – 1)

⇒ {a3 + (1/a3)}= (± 4) (13)

⇒ {a3 + (1/a3)}= (± 52)

15. If {m2+ (1/m2) = 51; find the value of m3– (1/m3)

m2 + (1/m2) = 51

We know that,

{m – (1/m)}2 = m2 + (1/m2) – 2

⇒ {m – (1/m)}2 = 51 – 2

⇒ {m – (1/m)}2 = 49

⇒ {m – (1/m)}2 = (7)2

⇒ {m – (1/m)} = 7

Cubing on both sides, we get,

{m – (1/m)}3 = (7)3

⇒ m3 – (1/m3) – 3 {m – (1/m)} = 343

⇒ m3 – (1/m3) – 3 (7) = 343

⇒ m3 – (1/m3) – 21 = 343

⇒ m3 – (1/m3) = 343 + 21

We get,

m3 – (1 / m3) = 364

16. If a2 – 3a – 1 = 0 and a ≠ 0, find ;

(i) a – 1/a,

(ii) a + 1/a,

(iii) a2 – 1/a2

(i)

(ii)
(iii)

17. If 2x + 3y = 10 xy = 5: find the value of 4x2 + 9y2

18. If x + y + z = 12 and xy + yz + zx = 27; find x2 + y2 + z2.

19. If a2 + b2 + c2 = 41 and a + b + c = 9; find ab + bc + ca.

20. If p2 + q2 + r2 = 82 and pq + qr + pr = 18; find p + q + r.

21. If x + y + z = p and xy + yz + zx = q; find x2 + y2 + z2

### Exercise 4.2

1. Find the cube of:

(i) 2a – 5b

(ii) 4x + 7y

(iii) 3a + 1/3a

(iv) 4p – 1/p

(v) 2m/3n + 3n/2m

(vi) a – 1/a + b

(i)

(ii)
(iii)
(iv)
(v)
(vi)

2. If 5x + 1/5x = 7; find the value of 125x3 + 1/125x3.

3. If 3x – 1/3x = 9: find the value of 27x3 – 1/27x3

4. If x + 1/x = 5, find the value of x2 + 1/x2, x3 + 1/x3 and x4 + 1/x4.

5. If a – 1/a = 7, find a2 + 1/a2, a2 – 1/a2 and a3 – 1/a3

6. If a2 + 1/a2 = 14; find the value of a + 1/a

(b) If a2 + 1/a2 = 14; find the value of a3 + 1/a3

(a)

(b)

7. If m2 + 1/m2 = 51; find the value of m3 – 1/m3

8. If 9a2 + 1/9a2 = 23; find the value of 27a3 + 1/27a3

9. If x2 + 1/x2 = 18; find :

(i) x – 1/x,

(ii) x3 – 1/x3

x2 + 1/x2 = 18

(i)

(ii)

10. If p + 1/p = 6; find :

(i) p2 + 1/p2,

(ii) p4 + 1/p4

(iii) p3 + 1/p3

(i)

(ii)

(iii)

11. If r – 1/r = 4; find

(i) r2 + 1/r2

(ii) r4 + 1/r4,

(iii) r3 – 1/r3

(i)

(ii)
(iii)

12. If a + 1/a = 2, then show that a2 + 1/a2 = a3 + 1/a3 = a4 + 1/a4

13. If x + 1/x = p, x – 1/x = q; find the relation between p and q.

14. If a + 1/a = p; then show that a3 + 1/a3 = p(p2 – 3)

15. If (a + 1/a)2 = 3; then show that a3 + 1/a3 = 0

16. If a + b + c = 0; then show that a3 + b3 + c3 = 3abc.

17. If a + 2b + c = 0; then show that a3 + 8b3 + c3 = 6abc

18. If x3 + y3 = 9 and x + y = 3, find xy.

19. If a + b = 5 and ab = 2, find a3 + b3

20. If p – q = - 1 and pq = - 12, find p3 – q3

21: If m – n = - 2 and m3 – n3 = - 26, find mn.

22: If 2a – 3b = 10 ab = 16; find the value of 8a3 – 27b3.

23: If x + 2y = 5, then show that x3 + 8y3 + 30xy = 125.

24. Simplify

(a) (4x + 5y)2 + (4x – 5y)2

(b) (7a + 5b)2 – (7a – 5b)2

(c) (a + b)3 = (a – b)3

(d) (a – 1/a)2 + (a + 1/a)2

(e) (x + y – z)2 + (x – y + z)2

(f) (a + 1/a)3 – (a – 1/a)3

(g) (2x + y)(4x2 – 2xy + y2)

(h) (x – 1/x)(x2 + 1 + 1/x2)

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

(j) (1 + x)(1 – x)(1 – x + x2)(1 + x + x2)

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

(l) (3x + 5y + 2z)(3x – 5y + 2z)

(m) (2x – 4y + 7)(2x + 4y + 7)

(n) (3a – 7b + 3)(3a – 7b + 5)

(o) (4m – 5n – 8)(4m – 5n + 5)

(a)

(b)

(c)
(d)
(e)

(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)

25: Evaluate the following:

(i) (3.29)3 + (6.71)3

(ii) (5.45)3 + (3.55)3

(iii) (8.12)3 – (3.12)3

(iv) 7.16×7.16 + 2.16×7.16 + 2.16×2.16

(v) 1.81×1.81 – 1.81×2.19 + 2.19×2.19