Example 1 Factorize the polynomial $f(x)=x^{8}+x^{7}+1$ over the integers.
Let $x=10$, then $f(10)=10^{8}+10^{7}+1=110000001=3 \cdot 37 \cdot 880991$.
Since $3, 37, 990991$ are three prime numbers, substituting $x=10$ back, we have $3=10-7=x-7, 37=30+7=40-3$, i.e., $37=3x+7$ or $43-x$. Noting Theorem 5, but the constant term of $f(x)$ is 1, so the constant term of the factor of $f(x)$ should also be 1. Therefore, $x-7, 3x+7$ or $4x-3$ cannot be the linear factors of $f(x)$.
Notice that $3 \cdot 37=111$, $f(x)$ may have a quadratic factor $x^{2}+x+1$. When 990991 is substituted with $x=10$,
$$
\text { we have } \begin{aligned}
990991 & =9 \cdot 10^{5}+9 \cdot 10^{4}+9 \cdot 10^{2}+9 \cdot 10+1 \\
& =(10-1) \cdot 10^{5}+(10-1) \cdot 10^{4}+(10-1) \cdot 10^{2}+(10-1) \cdot 10+1 \\
& =10^{6}-10^{4}+10^{3}-10+1 \\
& =x^{6}-x^{4}+x^{3}-x+1
\end{aligned}
$$

Upon verification, $x^{2}+x+1, x^{6}-x^{4}+x^{3}-x+1$ are indeed factors of $f(x)$. Therefore,
$$
f(x)=\left(x^{2}+x+1\right) \cdot\left(x^{6}-x^{4}+x^{3}-x+1\right) .
$$