Continuous practice using OP Malhotra Class 9 Solutions Chapter 3 Expansions Ex 3(B) can lead to a stronger grasp of mathematical concepts.

## S Chand Class 9 ICSE Maths Solutions Chapter 3 Expansions Ex 3(B)

Question 1.
Find the value of:
(i) a² + b² when a + b = 9, ab = 20
(ii) p² + q² if p – q = 6 and p + q = 14
(iii) mn if m + n = 8, m – n = 2
(iv) x² + $$\frac { 1 }{ x² }$$ and x4 + $$\frac{1}{x^4}$$ if x + $$\frac { 1 }{ x }$$ = 3
(v) x – $$\frac { 1 }{ x }$$ and x² – $$\frac { 1 }{ x² }$$ if x + $$\frac { 1 }{ x }$$ = $$\sqrt{5}$$
Solution:
(i) a + b = 9, ab – 20
= a² + b² = (a + b)² – 2ab
= (9)² – 2 x 20
= 81 – 40 = 41

(ii) p – q = 6, p + q = 14
2 (p² + q²) = (p + q)² + (p – q)²
= (14)² + (6)²
= 196 + 36 = 232
∴ p² + q² = $$\frac { 232 }{ 2 }$$ = 116

(iii) m + n = 8, m – n = 2
4mn = (m + n)² – (m – n)²
= (8)² – (2)² = 64 – 4 = 60
∴ mn = $$\frac { 60 }{ 4 }$$ = 15

(iv) x + $$\frac { 1 }{ x }$$ = 3
Squaring both sides,

(v) x + $$\frac { 1 }{ x }$$ = $$\sqrt{5}$$
Squaring both sides,

Question 2.
(i) a² + b² + c² if a + b + c = 17 and ab + bc + ca = 30.
(ii) ab + be + ca, a + b + c = 15 and a² + b² + c² = 77
(iii) a + b + c if a² + b² + c² = 50 and ab + bc + ca =47.
Solution:
(i) a + b + c = 17
Squaring both sides,
(a + b + c)² = (17)²
⇒ a² + b² + c² + 2 (ab + bc + ca) = 289
⇒ a²+ b² + c² + 2 x 30 = 289
⇒ a² + b² + c² + 60 = 289
⇒ a² + b² + c² = 289 – 60 = 229
∴ a² + b² + c² = 229

(ii) a + b + c = 15 and a² + b² + c² = 77
a + b + c = 15
Squaring both sides,
(a + b + c)² = (15)²
a² + b² + c² + 2 (a + bc + ca) = 225
⇒ 77 + 2 (ab + bc + ca) = 225
⇒ 77 + 2 (ab + bc + ca) = 225
⇒ 2 (a + bc + ca) = 225 – 77 = 148
⇒ ab + bc + ca = $$\frac { 148 }{ 2 }$$ = 74
∴ ab + bc + ca = 74

(iii) a² + b² + c² = 50, and ab + be + ca = 47
(a + b + c)² = a² + b² + c² + 2 (ab + bc + ca)
= 50 + 2 x 47 = 50 + 94
= 144 = (± 12)²
∴ a + b + c = ± 12

Question 3.
(i) 8x³ + 84x²y + 294xy² + 343y³ if x = 1, y = 2
(ii) 27x³ – 27x²y + 9xy2 – y³ if x = 2, y = 1
Solution:
(i) x = 1, y = 2
8x³ + 84x²y + 294xy² + 343y³
= (2x)³ + 3 (2x)² x 7y + 3 x 2x × (7y)² + (7y)³
= (2x + 7y)³ = (2 x 1 + 7 x 2)³
= (2 + 14)³ = (16)³ = 4096

(ii) x = 2, y = 1
27x³ – 27x²y + 9xy² – y³
= (3x)³ – 3 (3x)² y + 3 (3x)y² – (y)³
= (3x – y)² = (3 x 2 – 1)³
= (6 – 1)³ = (5)³ = 125

Question 4.
(i) a³ + b³ if a + b = 3 and ab = 2
(ii) x³ + $$\frac { 1 }{ x³ }$$ if (x + $$\frac { 1 }{ x }$$) = 3
(iii) x³ – $$\frac { 1 }{ x³ }$$ if x² + $$\frac { 1 }{ x² }$$ = 18
(iv) x³ + $$\frac{1}{125 x^3} \text { if } x^2+\frac{1}{25 x^2}=8 \frac{3}{5}$$
Solution:
(i) a + b = 3, ab = 2
a³ + b² = (a + b)³ – 3ab (a – b)
= (3)³ – 3 x 2 x 3
= 27 – 18 = 9

(ii) x + $$\frac { 1 }{ x }$$ = 3
x³ + $$\frac { 1 }{ x³ }$$ = (x + $$\frac { 1 }{ x }$$³ – 3(x + $$\frac { 1 }{ x }$$)
= (3)³ – 3 x 3 = 27 – 9 = 18

(iii) x² + $$\frac { 1 }{ x² }$$ = 18
Subtracting 2 from both sides

(iv) x² + $$\frac{1}{25 x^2}=8 \frac{3}{5}=\frac{43}{5}$$
(x)² + $$\frac{1}{(5 x)^2}$$ = $$\frac { 43 }{ 5 }$$
Completing the left hand sides, a complete
square adding 2 × x × $$\frac { 1 }{ 5x }$$ i.e., $$\frac { 2 }{ 5 }$$ both sides

Question 5.
Evaluate:
(i) 102 x 98
(ii) 1003² – 997²
(iii) (10)³ – (5)³ – (5)³
Solution:
(i) 10² x 98
= (100 + 2) (100-2) {∵ {a + b)(a – b) = a² – b²)}
= (100)² – (2)²
= 10000 – 4
= 9996

(ii) (1003)² – (997)²
= (1003 + 997) (1003 – 997) {∵ a² – b² = (a + b) (a – b)}
= 2000 x 6 = 12000

(iii) (10)³ – (5)³ – (5)³ = (10)³ + (- 5)³ + (- 5)³
Let 10 = a, – 5 = b, – 5 = c and a + b + c – 10 – 5 – 5 = 0
∵ a + b + c = 0. then a³ + b³ + c³ = 3abc
⇒ (10)³ – (5)³ – (5)³ = 3 x 10 x (- 5) (- 5) = 750