Solve the equation

$$
x^{4}-7 x^{3}+13 x^{2}-7 x+1=0
$$
Solution. Since $x=0$ is not a root of the equation, we perform an equivalent transformation by dividing by $x^{2}$. After rearranging:

$$
\begin{aligned}
\left(x^{2}+\frac{1}{x^{2}}\right)-7\left(x+\frac{1}{x}\right)+13 & =0 \\
\left(\left(x+\frac{1}{x}\right)^{2}-2\right)-7\left(x+\frac{1}{x}\right)+13 & =0 \\
\left(x+\frac{1}{x}\right)^{2}-7\left(x+\frac{1}{x}\right)+11 & =0
\end{aligned}
$$

By substituting $y=x+\frac{1}{x}$, we can transform this into $y^{2}-7 y+11=0$, whose solutions are $y_{1,2}=\frac{7 \pm \sqrt{5}}{2}$. Therefore, the solutions to the original equation are the solutions to the equations $x+\frac{1}{x}=y_{i}$, i.e., $x^{2}-y_{i} x+1=0$. Since $y_{i}>2$, each equation has two distinct real roots, namely

$$
\begin{array}{ll}
x_{1}=\frac{7+\sqrt{5}+\sqrt{38+14 \sqrt{5}}}{4}, & x_{2}=\frac{7+\sqrt{5}-\sqrt{38+14 \sqrt{5}}}{4}, \\
x_{3}=\frac{7-\sqrt{5}+\sqrt{38-14 \sqrt{5}}}{4}, & x_{4}=\frac{7-\sqrt{5}-\sqrt{38-14 \sqrt{5}}}{4}
\end{array}
$$

Remark. With a similar substitution, every even $2 n$-th degree equation can be reduced to an $n$-th degree equation and at most $n$ quadratic equations, where the coefficients of $x^{k}$ and $x^{2 n-k}$ are the same for all $k=0,1, \ldots, 2 n$. It is customary to call such equations reciprocal equations, and the corresponding polynomials reciprocal polynomials. The basis for this naming is that if a number $c$ is a root of such a polynomial, then $\frac{1}{c}$ is also a root, with the same multiplicity as $c$. The reduction to a half-degree equation generally relies on the fact that (for every positive integer $k$) $x^{k}+\frac{1}{x^{k}}$ can be written as a polynomial in $x+\frac{1}{x}$ (of degree $k$). This can be proven, for example, by induction on $k$: for $k=1$, this is trivial; assume it is true for all $k<m$. Then it is also true for $k=m$, since

$$
x^{m}+\frac{1}{x^{m}}=\left(x^{m-1}+\frac{1}{x^{m-1}}\right) \cdot\left(x+\frac{1}{x}\right)-\left(x^{m-2}+\frac{1}{x^{m-2}}\right)
$$

and by the induction hypothesis, $x^{m-1}+\frac{1}{x^{m-1}}$ and $x^{m-2}+\frac{1}{x^{m-2}}$ can be written as polynomials in $x+\frac{1}{x}$ (of the appropriate degree).