Given $a+b+c=0$, simplify the following. I am here again to ask a question about an exercise I saw around but i'm having a lot of trouble with. I know the answer is 3abc, but as in many of my questions, I am interested in the why and how.
Given $a+b+c=0$, simplify:
$$\frac{(a^3-abc)^3+(c^3-abc)^3+(b^3-abc)^3}{(c^2-ab)(b^2-ac)(a^2-bc)}$$

If you just sub in $c=-a-b$ then each factor of the denominator reduces to $(a^2+ab+b^2)$. 
Each term of the numerator has a factor that is one of these denominator factors as well. For instance the first term from the numerator is $a^3(a^2-bc)^3$, which is $a^3(a^2+ab+b^2)^3$.
So we have
$$\begin{align}
\frac{a^3(a^2+ab+b^2)^3+c^3(a^2+ab+b^2)^3+b^3(a^2+ab+b^2)^3}{(a^2+ab+b^2)^3}
\end{align}
$$
Now provided $a^2+ab+b^2\neq0$ (which is guaranteed the case if all our numbers are real), we have $$a^3+b^3+c^3$$
