Example 5. Factorize: $a^{4}+b^{4}+c^{4}+1-2\left(a^{2}\right.$ $\left.+b^{2}+c^{2}+a^{2} b^{2}+c^{2} a^{2}\right)+8 a b c$.
Solve, treating as a polynomial in $a$:
$$
\begin{array}{l}
f(a)=a^{4}-2\left(b^{2}+c^{2}+1\right) a^{2}+8 b c a \\
+\left(b^{4}+c^{4}+1-2 b^{2}-2 c^{2}-2 b^{2} c^{2}\right),
\end{array}
$$

where the last term, treated as a polynomial in $b$, is $F(b)=b^{4}$ $-2\left(1+c^{2}\right) b^{2}+\left(c^{4}+1-2 c^{2}\right) \cdot F(b)$, whose roots can be found as follows: using the formula, we get $b^{2}=1+c^{2} \pm 2 c$ $=(1 \pm c)^{2}$. Thus, the four roots of $F(b)$ are $1+c$, $1-c$, $-1+c$, and $-1-c$. Therefore, $F(b)=(b+c+1)(b+c-1)(b-c+1)(b-c-1)$. This way, $f(a)$ can be rewritten as:
$$
\begin{array}{r}
f(a)=a^{4}-2\left(b^{2}+c^{2}+1\right) a^{2}+8 b c a+ \\
(b+c+1)(b+c-1)(b-c+1)(b-c-1) .
\end{array}
$$

This expression can be factored into linear factors as in the previous example. We will use another method to directly find the quadratic factors.
Assume $f(a)$ can be factored into two quadratic factors:
$$
\begin{array}{l}
\left(a^{2}+p x+q\right) \quad\left(a^{2}+r x+s\right), \text { then } \\
p+r=0, \\
q r+s+q=-2\left(b^{2}+c^{2}+1\right), \\
n s+r q=8 l c, \\
q s=(1)+c+1)(b+c-1)(b-c+1) \\
\quad(b-c-1) .
\end{array}
$$

From (1), $p=-r$, substituting into (3), we get
$$
p(s-q) \times 8 b c \text {. }
$$

Therefore, $s-q$ should divide $8 l c$, and thus $s-q$ cannot contain terms with $b^{2}$, meaning that the terms with $b^{2}$ in $s$ and $q$ should be the same. From (4), we know that $s$ and $q$ are the product of two factors on the right side of (4).

Try $q= \pm(b-c+1)(b-c-1)$, $s= \pm(b+e+1)(b+c-1)$ (other combinations can also be tried), and find $s-q= \pm 4 b c$, $s+q=$ $\pm 2\left(b^{2}+c^{2}+1\right)$. Substituting into (5) and (1), we get $p= \pm 2$, $r=\mp 2$. Substituting into (2), we get $-4 \pm 2\left(b^{2}+c^{2}-\right.$ $1)=-2\left(b^{2}+c^{2}+1\right)$. It is clear that the " $\pm$ " should be negative, thus we get:
$$
\begin{aligned}
f(a)= & \left(a^{2}-2 a-(b-c+1)(b-c\right. \\
& -1))\left(a^{2}+2 a-(b+c+1)\right. \\
& \cdot(b+c-1))
\end{aligned}
$$

It is not difficult to further factor the two factors on the right side (for example, using the method of completing the square), and finally we get:
$$
\begin{aligned}
f(a)= & (a-b+c-1)(a+b-c-1) \\
& \cdot(a+b+c+1)(a-b-c+1) .
\end{aligned}
$$