<s>[INST] Solve degree-four polynomial equation $\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x$ How to solve 
$$\left(x^2-3x-5\right)^2-3\cdot \left(x^2-3x-5\right)-5=x?$$
I think there is a trick, because, if $X=x^2-3x-5$, I get 
$$ X^2-3X-5=x.$$
I don't know how to continue. 
I've also tried to factorize but it does not work. Any idea?
 [/INST] Factorize as follows,
$$(x^2-3x-5)^2-3(x^2-3x-5)^2-4=x+1$$
$$(x^2-3x-9)(x^2-3x-4)=x+1$$
$$(x^2-3x-9)(x-4)(x+1)-(x+1)=0$$
$$(x+1)(x^3-7x^2+3x+35)=0$$
$$(x+1)[(x^3-5x^2)-(2x^2-3x-35)]=0$$
$$(x+1)[x^2(x-5)-(2x+7)(x-5)]=0$$
$$(x+1)(x-5)(x^2-2x-7)=0$$
$$(x+1)(x-5)(x-1+2\sqrt2)(x-1-2\sqrt2)=0$$
Thus, the roots are $x=-1, 5, 1\pm 2\sqrt2$.
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