Solution:
It is known that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
Put $x+y+z=0$,
$x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x^{3}+y^{3}+z^{3}-3 x y z=0$
$x^{3}+y^{3}+z^{3}=3 x y z$
It is known that,
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
Put $x+y+z=0$,
$x^{3}+y^{3}+z^{3}-3 x y z=(0)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x^{3}+y^{3}+z^{3}-3 x y z=0$
$x^{3}+y^{3}+z^{3}=3 x y z$
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