# 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!

• While it's only a hint and not a solution, it is always not a bad idea to make a graph (DESMOS) just to see what is happening Commented Jul 1, 2020 at 21:57

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.