Question

Given x+y+z=0 then prove that x³+y³+z³=3xyz

Answer 1

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-xz)
=(0)(x²+y²+z²-xy-yz-xz) [given x+y+z=0]
=0
=+3xyz [Transposing].