Solve the equation $x^{4}-6 x^{3}+10 x^{2}-4 x=0$.

List 8
Solution: The original equation can be transformed into
$$
\left(x^{2}-3 x+2\right)^{2}-3\left(x^{2}-3 x+2\right)+2-x=0 \text {. }
$$

Let $F(x)=x^{2}-3 x+2$, then the two fixed points of $F(x)$ are $2+\sqrt{2}$ and $2-\sqrt{2}$ (by the definition of a function's fixed point, we get $x^{2}-3 x+2=x$, i.e., $x^{2}-4 x+2=0$, solving this yields).

From the conclusion of Example 7, we know that $2+\sqrt{2}$ and $2-\sqrt{2}$ are also two fixed points of $F(F(x))$. The other two fixed points of $F(F(x))$ can be obtained from the equation
$$
1+[1+(x-2+\sqrt{2})][1+(x-2-\sqrt{2})]=0,
$$

i.e., $x^{2}-2 x=0$, solving this yields $x_{1}=0, x_{2}=2$.
Therefore, the 4 roots of the equation $F(F(x))-x=0$ are $2+\sqrt{2}, 2-\sqrt{2}, 0,2$, which are the solutions to the original equation.

From this, we can see that for the algebraic equation $G(x)=0$, it can be converted into the equivalent equation $F(x)=x$, and by finding the fixed points of $F(x)$, we can obtain the roots of $G(x)=0$.