# Write the following cubes in expanded form:

i) (2x +1)³ ii) (2a - 3b)³ iii) (3x/2 + 1)³

iv) (x - 2y/3)³

**Solution:**

We will be using the below algebraic identities:

(x + y)³ = x³ + y³ + 3xy(x + y)

(x - y)³ = x³ - y³ - 3xy(x - y)

i) (2x + 1)³

Identity: (x + y)³ = x³ + y³ + 3xy(x + y)

Here x = 2x, y = 1

(2x +1)³ = (2x)³ + (1)³ + 3(2x)(1)(2x + 1)

= 8x³ + 1 + 6x(2x + 1)

= 8x³ + 1 + 12x² + 6x

= 8x³ + 12x² + 6x +1

ii) (2a - 3b)³

Identity: (x - y)³ = x³ - y³ - 3xy(x - y)

Here x = 2a, y = 3b

(2a - 3b)³ = (2a)³ - (3b)³ - 3(2a)(3b)(2a - 3b)

= 8a³ - 27b^{3} - 18ab(2a - 3b)

= 8a³ - 27b³ - 36a²b + 54ab²

= 8a³ - 36a²b + 54ab² - 27b³

iii) (3x/2 + 1)³

Identity: (x + y)³ = x³ + y³ + 3xy(x + y)

Here, x = 3x/2, y = 1

(3x/2 + 1)³ = (3x/2)³ + (1)³ + 3(3x/2)(1)(3x/2 + 1)

= (27/8)x³ + 1 + (9/2)x (3x/2 + 1)

= (27/8)x³ + 1 + (27/4)x² + (9/2)x

= (27/8)x³ + (27/4)x² + (9/2)x + 1

iv) (x - 2y/3)³

Identity: (x - y)³ = x³ - y³ - 3xy(x - y)

Here, x = x, y = 2y/3

(x - 2y/3)³ = x³ - (2y/3)³ - 3 (x)(2y/3)(x - 2y/3)

= x³ - (8/27)y³ - 2xy(x - 2y/3)

= x³ - (8/27)y³ - 2x²y + (4/3)xy²

= x³ - 2x²y + (4/3)xy² - (8/27)y³

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Write the following cubes in expanded form: i) (2x +1)³ ii) (2a - 3b)³ iii) (3x/2 + 1)³ iv) (x - 2y/3)³

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 6

**Summary:**

The expanded form of the cubes (2x +1)³, (2a - 3b)³, (3x/2 + 1)³ and (x - 2y/3)³ are 8x³ + 12x² + 6x +1, 8a³ - 36a²b + 54ab² - 27b³, (27/8)x³ + (27/4)x² + (9/2)x + 1, and x³ - 2x²y + (4/3)xy² - (8/27)y³ respectively.

**☛ Related Questions:**

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- Factorise: (i) 4x 2 + 9y 2 + 16z 2 + 12xy – 24yz – 16xz (ii) 2x 2 + y 2 + 8z 2 – 2 2 xy + 4 2 yz – 8xz

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