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Question -

Write the followingcubes in expanded form:

(i) (2x+1)3                          (ii) (2aтИТ3b)3

(iii) ((3/2)x+1)3                (iv) (xтИТ(2/3)y)3



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)= (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)

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