# Maharashtra Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.2

## Maharashtra State Board Class 8 Maths Solutions Chapter 5 Expansion Formulae Practice Set 5.2

Question 1.
Expand:
i. (k + 4)³
ii. (7x + 8y)³
iii. (7x + m)³
iv. (52)³
v. (101)³
vi. $$\left(x+\frac{1}{x}\right)^{3}$$
vii. $$\left(2 m+\frac{1}{5}\right)^{3}$$
viii. $$\left(\frac{5 x}{y}+\frac{y}{5 x}\right)^{3}$$
Solution:
i. Here, a = k and b = 4
(k + 4)³ = (k)³ + 3(k)² (4) + 3(k)(4)² + (4)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= k³ + 12k² + 3(k)(16) + 64
= k³ + 12k² + 48k + 64

ii. Here, a = 7x and b = 8y
(7x + 8y)³
= (7x)³ + 3(7x)² (8y) + 3(7x) (8y)² + (8y)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343x³ + 3(49x²)(8y) + 3(7x)(64y²) + 512y³
= 343x³ + 1176x²y + 1344xy² + 512y³

iii. Here, a = 7 and b = m
(7 + m)³ = (7)³ + 3(7)²(m) + 3(7)(m)² + (m)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 343 + 3(49)(m) + 3(7)(m²) + m³
= 343 + 147m + 21m² + m³

iv. (52)³ = (50 + 3)³
Here, a = 50 and b = 2
(52)³ = (50)³ + 3(50)² (2) + 3(50)(2)² + (2)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 125000 + 3(2500)(2) + 3(50)(4) + 8
= 125000 + 15000 + 600 + 8
=140608

v. (101)³ = (100 + 1)³
Here, a = 100 and b = 1
(101)³
= (100)³ + 3(100)²(1) + 3(100)(1)² + (1)³
…[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= 1000000 + 3(10000) + 3(100) (1) + 1
= 1000000 + 30000 + 300 + 1
= 1030301

vi. Here, a = x and b = $$\frac { 1 }{ x }$$

vii. Here, a = 2m and b = $$\frac { 1 }{ 5 }$$

viii. Here, a = $$\frac { 5x }{ y }$$ and b = $$\frac { y }{ 5x }$$