<s>[INST] How to evaluate $x^6+x^4+x^3+x^2+1=0$? There is a hint in the question, use the factorization of $x^5+x+1$.
 [/INST] Divide by $x^3$ and remember that
$$
x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
$$
so the equation becomes
$$
\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)
+\left(x+\frac{1}{x}\right)+1=0
$$
Set $t=x+1/x$ and you get
$$
t^3-2t+1=0
$$
Can you go on? Note that $t=1$ is a solution.
</s>