Three, (10 points) Given the multiplication formulas:
$$
\begin{array}{l}
a^{5}+b^{5}=(a+b)\left(a^{4}-a^{3} b+a^{2} b^{2}-a b^{3}+b^{4}\right) . \\
a^{5}-b^{5}=(a-b)\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) .
\end{array}
$$

Using or not using the above formulas, factorize the following expression:
$$
x^{8}+x^{6}+x^{4}+x^{2}+1
$$
$$
\begin{array}{l}
x^{8}+x^{6}+x^{4}+x^{2}+1 \\
=\frac{x^{10}-1}{x^{2}-1}=\frac{x^{5}-1}{x-1} \cdot \frac{x^{5}+1}{x+1} \\
=\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right) .
\end{array}
$$

The application of the formula is as follows:
$$
\begin{array}{l}
x^{8}+x^{6}+x^{4}+x^{2}+1 \\
=\frac{x^{10}-1}{x^{2}-1}=\frac{x^{5}-1}{x-1} \cdot \frac{x^{5}+1}{x+1} \\
=\left(x^{4}+x^{3}+x^{2}+x+1\right)\left(x^{4}-x^{3}+x^{2}-x+1\right) .
\end{array}
$$