Question

If x+y+z=6 and xy+yz+zx=12 then show that x^3+y^3+z^3=3xyz

Answer 1

x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)
=6(12-12)
=0
so x+y+z=0
then x^3+y^3+z^3=3xyz