Example 3. Factorize: $x^{12}+x^{9}+x^{6}+x^{3}+1$.
Let $\omega$ be a fifth root of unity other than 1. Substituting $x = \omega$ into the polynomial, we get
$$
\begin{array}{l}
\omega^{12}+\omega^{9}+\omega^{6}+\omega^{3}+1 \\
=\left(\omega^{5}\right)^{2} \omega^{2}+\omega^{5} \omega^{4}+\omega^{5} \omega^{1}+\omega^{3}+1 \\
=\omega^{2}+\omega^{4}+\omega^{1}+\omega^{3}+1 \\
=\frac{\omega^{5}-1}{\omega-1}=0 .
\end{array}
$$

By Theorem 1, the original polynomial contains the factor $x^{4}+x^{3}+x^{2}+x+1$. Using polynomial division, it is easy to obtain the quotient polynomial, then
$$
\begin{array}{l}
x^{12}+x^{9}+x^{6}+x^{3}+1 \\
=\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{8}-x^{7}+x^{5}-x^{4}\right. \\
\left.\quad+x^{3}-x+1\right) .
\end{array}
$$