If x+y+z=0, show that x^3+y^3+z^3=3xyz

We know that,

x³ + y³ + z³ – 3xyz = (x + y + z) (x² + y² + z² – xy – yz – xz)

Now, put (x + y + z) = 0

x³ + y³ + z³ – 3xyz = (0) (x² + y² + z² – xy – yz – xz)

x³ + y³ + z³ – 3xyz = 0