9・118 (1) Factorize $x^{12}+x^{9}+x^{6}+x^{3}+1$.
(2) For any real number $\theta$, prove: $5+8 \cos \theta+4 \cos 2 \theta+\cos 3 \theta \geqslant 0$.
[Solution] (1) Since $x^{15}-1=\left(x^{3}-1\right)\left(x^{12}+x^{9}+x^{6}+x^{3}+1\right)$,
$$x^{15}-1=\left(x^{5}-1\right)\left(x^{10}+x^{5}+1\right)$$

Therefore, $x^{12}+x^{9}+x^{6}+x^{3}+1=\frac{\left(x^{4}+x^{3}+x^{2}+1\right)\left(x^{10}+x^{5}+1\right)}{x^{2}+x+1}$.
Noting that if $x^{2}+x+1=0$ it implies $x^{10}+x^{5}+1=0$, thus $x^{10}+x^{5}+1$ must be divisible by $x^{2}+x+1$. Using polynomial division, we easily find that
$$x^{10}+x^{5}+1=\left(x^{2}+x+1\right)\left(x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1\right),$$

Therefore, $x^{12}+x^{9}+x^{6}+x^{3}+1=\left(x^{4}+x^{3}+x^{2}+1\right)\left(x^{8}-x^{7}+x^{5}-\right.$
$$\left.x^{4}+x^{3}-x+1\right)$$
(2)
$$\begin{aligned}
& 5+8 \cos \theta+4 \cos 2 \theta+\cos 3 \theta \\
= & 5+8 \cos \theta+4\left(2 \cos ^{2} \theta-1\right)+\left(4 \cos ^{3} \theta-3 \cos \theta\right) \\
= & 1+5 \cos \theta+8 \cos ^{2} \theta+4 \cos ^{3} \theta \\
= & 1+\cos \theta+4 \cos \theta(1+\cos \theta)^{2} \\
= & (1+\cos \theta)(2 \cos \theta+1)^{2} \geqslant 0
\end{aligned}$$