ABC's problem

a+b = -c

Cubing on both sides

$a^3+b^3+3ab(a+b)$ = - $c^3$

So $a^3+b^3+3ab(-c)$ + $c^3$ = 0.

So $a^3+b^3+c^3$ = 3abc.

We know that $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

So if $a+b+c=0$ We have $a^3+b^3+c^3-3abc=0$

$a^3+b^3+c^3=3abc$

$\frac{a^3+b^3+c^3}{abc}=3$