Q5 of 59 Page 7

Expand :

(i) (3a + 5b)3


(ii) (4x – 3y)3


(iii)

(i) (3a + 5b)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(3a + 5b)3


= (3a)3 + 3 (3a)2(5b) + 3(3a)(5b)2 + (5b)3


= 27a3 + 135a2b + 225ab2 + 125b3


(ii) (4x – 3y)3


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


(4x – 3y)3


= (4x)3 + 3(4x)2(– 3y) + 3 × 4x (– 3y)2 + (– 3y)3


= 64x3 – 192 x2 + 108y2 – 27y3


(iii)


Here the identity used is


(a + b)3 = a3 + 3a2b + 3ab2 + b3


=


=


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