Example 4. Factorize $x y z\left(x^{3}+y^{3}+z^{3}\right)-x^{3} y^{3}$ $-y^{3} z^{3}-z^{3} x^{3}$.
Solve: The original expression can be seen as a polynomial in $x$
$$
\begin{aligned}
f(x)= & x^{4} y z-x^{3}\left(y^{3}+z^{3}\right)-x y z\left(y^{3}\right. \\
& \left.+z^{3}\right)-y^{3} z^{3} .
\end{aligned}
$$

If $f(x)$ has a linear factor $a x+b$, then $a$ divides $y z$, and $b$ divides $y^{3} z^{3}$. Therefore, we only need to consider the following four possibilities:
(1) $x \pm y^{n} z^{m}$;
(2) $y x \pm y^{n} z^{m}$;
(3) $z x \pm y^{n} z^{m}$;
(4) $y z x \pm y^{n} z^{m}$.
Here, $n, m$ can take the values $0, 1, 2, 3$ (how many combinations are there?).
There are various ways to eliminate unnecessary combinations. For example, since $y, z$ are not factors of $f(x)$, the terms $y x \pm y z^{m}, y x \pm y^{2} z^{m}, y x \pm y^{3} z^{m}$ in (2) are also not factors of $f(x)$. Therefore, in (2) we only need to consider $y x \pm z^{m}$, in (3) only $z x \pm y^{n}$, and in (4) only $y z x \pm 1$.
We also need to consider symmetry. If (is a factor of $f(x)$, then $2 x+y^{\prime}$ in (3) should also be, and vice versa. For example, we only need to consider (2), not (3). Also, if $y x+z^{3}$ in (2) is a factor of the original expression, then $z y+x^{3}, x z+y^{3}$ should also be factors. In this case, $\left(y x+z^{3}\right)\left(z y+x^{3}\right)\left(x z+y^{3}\right)$ should divide $f(x)$, but this is impossible (why?). Using this perspective, we can eliminate a large number of cases.
Thus, we are left with the following possibilities:
(1) $x \pm 1, x \pm y, x \pm y z$;
(2) $y x \pm 1, y x \pm z^{2}$;
(4) $y z x \pm 1$.
Now, we find $f(-y)=y^{8}-y^{2} z^{4}=y^{2}\left(y^{2}+z^{2}\right)(y+z)(y-z)$.
By evaluating $g(-y)$ for each of the expressions in (1), (2), and (4), and observing whether they can divide the above expression, we can narrow down the possibilities to:
(1) $x-y$;
(2) $y x \pm z^{2}$.

At this point, it is not difficult to use trial division to find the factor $y x-z^{2}$, and we finally get:
$$
f(x)=\left(x^{2}-y z\right)\left(y^{2}-z x\right)\left(z^{2}-x y\right) .
$$