Question

Find x^3+y^3-12xy+64 ,when x+y= -4

Answer 1

since it is given that x+y=-4 by cubing on both sides we get
x^3+y^3+3xy(x+y) = -64
x^3+y^3+3xy(-4) = -64
x^3+y^3-12xy = -64
hence substituting this we get -64+64 = 0
hence the answer is 0

Answer 2

= x^3 + y^3  - 12xy + 64

= x^3 + y^3 + 4^3 - 3 * x * y * 4

We know that,  a^3 + b^3 + c^3 - 3abc = ( a + b+c )(a^2+b^2+c^2-ab-bc-ca)

= ( x + y + 4 ) ( x^2 + y^2 + 4^2 - xy - 4y - 4x) --------------- eq.1

We have, ( x + y ) =-4, by putting this value in eq.1

= ( -4 + 4 ) ( x^2 + y^2 + 16 -xy - 4y - 4x )

= 0 ( x^2 + y^2 + 16 - xy -4y -4x )

= 0.