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:
$$\frac{(a^3-abc)^3+(c^3-abc)^3+(b^3-abc)^3}{(c^2-ab)(b^2-ac)(a^2-bc)}$$
Your idea that $a^3+b^3+c^3=3abc$ is a good one.
Consider your numerator $x^3+y^3+z^3$, where $x = a^3-abc$ and so forth, try to see what $x+y+z$ is, and what the relationship between $x^3+y^3+z^3$ and $3xyz$ is.
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 $$a^3+b^3+c^3$$
You can factor e.g. $a^3 - a b c = a(a^2 - b c)$, which matches a factor in the denominator. This suggests a sum of fractions.
You can always write out the numerator and denominator. The similarity of the terms gives hope that there will be lots of cancellation.
Try replacing simple values, like $a = b = c$ and see what happens.
