If $a^3+b^3+c^3=3abc$, then prove that $a=b=c$.

$\displaystyle \text{By the identity, }a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2+ab+bc+ca)\text{ we find either }a+b+c=0\text{ or }a^2+b^2+c^2+ab+bc+ca=0.\text{ the first thing is false since we arrive at a contradiction.}$

$\displaystyle \text{ I tried many things even by expressing }\begin{aligned} a^2+b^2+c^2+ab+bc+ca&=a^2+b^2+c^2+2ab+2bc+ca-ab-bc\\&=(a+b)^2+(b+c)^2-ab-bc+ca\end{aligned}$

I see nothing helpful. Is the identity I am using is right?