Question -
Answer -
(i) (2x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
(2x+1)3= (2x)3+13+(3├Ч2x├Ч1)(2x+1)
= 8x3+1+6x(2x+1)
= 8x3+12x2+6x+1
(ii) (2aтИТ3b)3
Using identity,(xтАУy)3 = x3тАУy3тАУ3xy(xтАУy)
(2aтИТ3b)3 = (2a)3тИТ(3b)3тАУ(3├Ч2a├Ч3b)(2aтАУ3b)
= 8a3тАУ27b3тАУ18ab(2aтАУ3b)
= 8a3тАУ27b3тАУ36a2b+54ab2
(iii) ((3/2)x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
((3/2)x+1)3=((3/2)x)3+13+(3├Ч(3/2)x├Ч1)((3/2)x+1)
(iv) (xтИТ(2/3)y)3
Using identity, (x тАУy)3 = x3тАУy3тАУ3xy(xтАУy)