jmc

Let $y=x^2-13x-8.$ Then we can write the given equation as $$\frac{1}{y+24x}+\frac{1}{y+15x}+\frac{1}{y}=0.$$Multiplying everything by $(y+24x)(y+15x)y,$ we get $$(y+15x)y+y(y+24x)+(y+24x)(y+15x)=0.$$This simplifies to $360x^2+78xy+3y^2=0,$ which factors as $3(20x+y)(6x+y)=0.$ Hence, $20x+y=0$ or $6x+y=0.$

If $20x+y=0,$ then $20x+x^2-13x-8=x^2+7x-8=(x-1)(x+8)=0,$ so $x=1$ or $x=-8.$

If $6x+y=0,$ then $6x+x^2-13x-8=x^2-7x-8=(x-8)(x+1)=0,$ so $x=8$ or $x=-1.$ Thus, the solutions are $\boxed{8,1,-1,-8}.$