Solve the equation $\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$ Problem
Solve the equation $$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$$
What I've tried
First I tried factoring the denominators but only the second one can be factored as $(x+4)(x-2)$.
Then I tried substituting $y = x^2 - 8$ but that didn't lead me anywhere.
Where I'm stuck
I don't know how to start this problem. Any hints?
P.S. I would really appreciate it if you give me hints or at least hide the solution. Thanks for all your help in advance!

You started very well. To make things easier, set $A=x^2+7x-8$ (*) and the equation rewrites
$$\frac{1}{A+4x} + \frac{1}{A-5x} + \frac{1}{A-20x} = 0.$$
Denominators are not allowed to be zeros. We solve
$$(A-5x)(A-20x)+(A+4x)(A-20x)+(A+4x)(A-5x)=0$$
or equivalently
$$3A^2-42Ax=0,$$ or even
$$3(x^2+7x-8)(x^2-7x-8)=0,$$
which is easy to finish.
(*) I noticed that $x=1$ satisfies, and decided to make profit from it.
