Question

if x +y+z=0, show that x cube + y cube + z cube = 3xyz

Answer 1

x+y=-z
cubing both side
x cube plus y cube plus 3xy(x+y)=-z cube
x cube plus y cube plus 3xy (-z) = -z cube
x cube plus y cube plus z cube - 3xyz=0
x cube plus y cube plus z cube=3xyz

Answer 2

We know that , x cube + y cube + z cube -3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
Putting x + y + z = 0 ,
x^3+y^3+z^3-3xyz = (0)(x^2+y^2+z^2-xy-yz-zx)
x^3+y^3+z^3-3xyz = 0
x^3+y^3+z^3 = 0 + 3xyz = 3xyz
x^3+y^3+z^3 = 3xyz
Hence proved