amanvermagmat wrote:

This can also be solved by plugging in values. Lets assume x = y = 1.

So 4(xy)^3 + (x^3 − y^3)^2 = 4 *(1*1)^2 + (1-1)^2 = 4.

Thus we have to look for an option where when we put x=1 and y=1, we should get our answer as '4'.

Check A) 1-1 = 0

B) (1+1)^3 = 8

C) (1+1)^3 = 8

D) (1-1)^2 = 0

E) (1+1)^2 = 4

Only E satisfies. Thus E answer

4x^3*y^3 + x^6 - 2(x^3- y^3) + y^6. On opening up the brackets we get, 4x^3*y^3 + x^6 - 2x^3y^3 + y^6. Cancel out 4x^3y^3 and 2x^3y^3. Finally we get, x^6 + 2x^3y^3 +y^6. Therefore, the answer is E.