11.1. Solve the equation $x^{10}-3 x^{4}+x^{2}+1=0$.
Answer: $x= \pm 1$. Solution. Making the substitution $t=x^{2}, t \geq 0$, we get the equation $t^{5}-3 t^{2}+t+1=0$. Noticing that $t=1$ is a root and dividing the left side by $(t-1)$, we will have $(t-1)\left(t^{4}+t^{3}+t^{2}-2 t-1\right)=0$. The polynomial $t^{4}+t^{3}+t^{2}-2 t-1$ also has a root $t=1$. After dividing this polynomial by $(t-1)$, we get the equation $(t-1)^{2}\left(t^{3}+2 t^{2}+3 t+1\right)=0$. The polynomial $t^{3}+2 t^{2}+3 t+1$ has only positive coefficients and therefore has no non-negative roots. Thus, $t=1$ is the only root (of multiplicity 2) and, returning to the variable $x$, we get two roots $x= \pm 1$.