Example 2 Factorize: $f(x)=x^{8}+x^{4}+1$.

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Solve $f(x)$ has no linear factors (why?) Let's examine the quadratic factors. This polynomial is of degree 8, and using method 1 would be quite difficult. We adopt method 2. Suppose $g(x)$ is a quadratic factor. Then $g(0)$ divides $f(0)$ $=1, g( \pm 1)$ divides $f( \pm 1)=3$. From $g(1)$ $+g(-1)=2(1+g(0))$, we know only the following quadratic polynomials need to be examined: (see the table on the left)
Since $f( \pm 2)=273=3 \times 7 \times 11$, we only need to examine the last two cases. After trial division, they are both factors of $f(x)$, and:
$$
\begin{aligned}
f(x)= & \left(x^{2}+x+1\right)\left(x^{2}-x+1\right) \\
& \cdot\left(x^{4}-x^{2}+1\right) .
\end{aligned}
$$

Therefore, $f(x)$ has no other quadratic factors, and they cannot divide $x^{4}-x^{2}+1$, so $f(x)$ cannot be further factored.