Example 8. Solve the equation:
$$
6 x^{4}-13 x^{3}+12 x^{2}-13 x+6=0 .
$$
Since $x=0$ is not a root of the original equation, we divide both sides of the equation by $x^{2}$, getting
$$
\begin{array}{l}
6 x^{2}-13 x+12-\frac{13}{x}+\frac{6}{x^{2}}=0, \\
6\left(x^{2}+\frac{1}{x^{2}}\right)-13\left(x+\frac{1}{x}\right)+12=0 .
\end{array}
$$

Let $y=x+\frac{1}{x}$, then $x^{2}+\frac{1}{x^{2}}=y^{2}-2$.
Substituting into the above equation, we get $6\left(y^{2}-2\right)-13 y+12=0$.
Solving this equation, we get $y_{1}=0, y_{2}=\frac{13}{6}$.
The equation $x+\frac{1}{x}=0$ has no real solutions.
From the equation $x+\frac{1}{x}=\frac{13}{6}$, the solutions to the original equation are $x_{1}=\frac{2}{3}, \quad x_{2}=\frac{3}{2}$.