Example 6. Solve the equation: $1+2 x+3 x^{2}+4 x^{3}+5 x^{4}+$ $6 x^{5}+5 x^{6}+4 x^{7}+3 x^{8}+2 x^{9}+x^{10}=0$.
From Example 4, we know that the equation is equivalent to:
$$
\begin{array}{l}
(x+1)^{2}\left(1-x+x^{2}\right)^{2}\left(1+x+x^{2}\right)^{2}=0 \\
\Leftrightarrow(x+1)^{2}=0 \text { or }\left(1-x+x^{2}\right)^{2}=0 \text { or } \\
\left(1+x+x^{2}\right)^{2}=0 .
\end{array}
$$

Therefore, the roots of the original equation are $x=-1$ or $x=$ $\frac{1 \pm \sqrt{3} i}{2}$ or $x=\frac{-1 \pm \sqrt{3} i}{2}$, and these roots are all double roots.