Q.6. Write the following cubes in expanded form:

(i) (2x + 1)3 (ii) (2a - 3b)3 (iii) [32 x + 1]3 (iv) [x - 23 y]3

Solution:

(i) Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)

We can write,

(2x + 1)= (2x)3 + 13 + (3×2x×1)(2x + 1)

= 8x3 + 1 + 6x(2x + 1)

= 8x3 + 12x2 + 6x + 1

(ii) Using identity, (x - y)3 = x3 - y3 - 3xy(x - y)

We can write,

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

= 8a3 - 27b3 - 18ab(2a - 3b)

= 8a3 - 27b3 - 36a2b + 54ab2

(iii) Using identity, (a + b)3 = a3 + b3 + 3ab(a + b)

We can write,

32x + 13= 32x3+13+ 3×32x×132x + 1

=278x3 + 1 +92x32x + 1

=278x3 + 1 +274x2 +92x

=278x3+274x2 +92x + 1

(iv) Using identity, (a - b)3 = a3 - b3 - 3ab(a -b)

We can write,

x-23y 3= x3-23y3+ 3×23y×xx-23y

= x3 -827y3- 2xyx -23y

= x3 -827y3- 2x2y +43xy2