jmc

From the equation $a+b+c=0,$ $c=-a-b.$ Hence, $$\begin{array}{rl}\frac{a^3+b^3+c^3}{abc}& =-\frac{a^3+b^3-(a+b{)}^3}{ab(a+b)}\\ & =\frac{3a^{2}b+3a{b}^2}{ab(a+b)}\\ & =\frac{3ab(a+b)}{ab(a+b)}\\ & =\boxed{3}.\end{array}$$By the Multivariable Factor Theorem, this implies that $a+b+c$ is a factor of $a^3+b^3+c^3-3abc.$ We can then factor, to get the factorization. $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc).$$