Question

If (x+y+z)=0 then prove that( x cube + y cube +z cube) =3xyz

Answer 1

x³+y³+z³-3xyz=(x+y+z)( x²+y²+z²-xy-yz-zx)
x³+y³+z³-3xyz=(0)(x²+y²+z²-xy-yz-zx)
x³+y³+z³-3xyz=0
x³+y³+z³=3xyz