If $a+b+c = 0$,

When $a+b+c=0$, the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ has a zero right side, so $a^3+b^3+c^3-3abc=0$. Therefore $a^3+b^3+c^3=3abc$.