## 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 = \(\frac { y }{ 5x }\)