Problem 531

3. Let $x^{2}-10 x-49=t$, then the original equation can be transformed into $\frac{1}{t+20}+\frac{1}{t+4}-\frac{2}{t-20}=0$. Clearing the denominators, we get
$$(t+4)(t-20)+(t+20)(t-20)-2(t+20)(t+4)=0 .$$

Expanding and simplifying, we get $-64 t=640$, i.e., $t=-10$. Substituting $t=-10$ yields $x^{2}-10 x-49=-10$. Solving this, we get $x_{1}=13$, $x_{2}=-3$.
Upon verification, $x_{1}=13, x_{2}=-3$ are both roots of the original equation.