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

If x+y+z=0, show that x3+y3+z3=3xyz

If x+y+z=0, show that x3+y3+z3=3xyz
Given that :
x + y + z = 0 …(1)
We know that,
x3+ y3 + z3 - 3xyz =(x + y + z)(x2 + y2 + z2 - xy - yz - zx) …(2)
Substituting the value of (1) in (2), we get
x3 + y3 + z3 - 3xyz = 0
x3 + y3 + z3 = 3xyz