Example 2 Factorize $f(x, y, z)=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$.

 

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Example 2 Factorize $f(x, y, z)=x^{3}(y-z)+y^{3}(z-x)+z^{3}(x-y)$.
Since $f(x, y, z)$ is a symmetric function in $x, y, z$, it must have the factors $x-y, y-z, z-x$, thus it has the factor $(x-y)(y-z)(z-x)$. Also, since $f(x, y, z)$ is a homogeneous polynomial of degree four, it must have another linear symmetric factor, so we can assume
$$
f(x, y, z)=(x-y)(y-z)(z-x) \cdot k(x+y+z).
$$

By comparing the coefficient of the term $x^{3} y$ in the original expression with the assumed form, we get $k=-1$.
Therefore, $f(x, y, z)=-(x-y)(y-z)(z-x)(x+y+z)$.
Note: This example can also be solved using calculus to factorize.
Let the given expression be $g(x)$, treating $y, z$ as constants, and differentiate with respect to $x$, we have
$$
g^{\prime}(x)=3 x^{2}(y-z)-y^{3}+z^{3}.
$$

Integrating with respect to $x$ gives $g(x)=x^{3}(y-z)-\left(y^{3}-z^{3}\right) x+c_{1}$, and $c_{1}=g(0)=y^{3} z-z^{3} y$, thus $g(x)=x^{3}(y-z)-\left(y^{3}-z^{3}\right) x+y z\left(y^{2}-z^{2}\right)=(y-x) \cdot h(x)$.
Here, $h(x)=x^{3}-x y^{2}-x y z-x z^{2}+y^{2} z+y z^{2}$.
Differentiating with respect to $y$, we get $h^{\prime}(y)=-2 x y-x z+2 y z+z^{2}=2 y(z-x)-x z+z^{2}$.
Integrating with respect to $y$ again, we get $h(y)=y^{2}(z-x)+z(z-x) y+c_{2}$, and $c_{2}=h(0)=x^{3}-x z^{2}$. Thus, $h(x)=(z-x)\left(y^{2}+y z-x z-x^{2}\right)=(z-x)(y-x)(x+y+z)$.
Therefore, $f(x)=-(x+y+z)(x-y)(y-z)(z-x)$.