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since we know that
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
and(x+y+z)=0
then,x^3+y^3+z^3-3xyz=0
thereforex^3+y^3+z^3=3xyz
x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
and(x+y+z)=0
then,x^3+y^3+z^3-3xyz=0
thereforex^3+y^3+z^3=3xyz
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