Example 4. Factorize: $x^{10}+x^{5}+1$.
Solve: Substituting $x=\omega$ into $x^{10}+x^{5}+1$, we get $\omega^{10}+\omega^{5}+1=\omega+\omega^{2}+1=0$.
$\therefore x^{10}+x^{5}+1$ must have a factor $x^{2}+x+1$.
By synthetic division, we get $\left(x^{10}+x^{5}+1\right)$
$$
\begin{array}{c}
\div\left(x^{2}+x+1\right) \\
=x^{8}-x^{7}+x^{6}-x^{4}+x^{3}-x+1 \\
\therefore x^{10}+x^{6}+1=\left(x^{2}+x+1\right) \\
\cdot\left(x^{8}-x^{7}+x^{6}-x^{4}+x^{3}-x+1\right)
\end{array}
$$