Example 9. Factorize $a b(a-b)+b c(b-c)+c a(c-a)$.
Replace $a$ with $b$, the original expression $=b^{2}(b-b)+b c(b-c)+c b$ $(c-b)=0$, so the original expression has a factor $(a-b)$;

Replace $c$ with $b$, the original expression $=a c(a-c)+c^{2}(c-c)+c a(c-$ a) $=0$, so the original expression has a factor $(b-c)$;

Replace $c$ with $a$, the original expression $=a b(a-b)+b a(b-a)+a^{2}(a-$ a) $=0$, so the original expression has a factor $(c-a)$.

Assume $a b(a-b)+b c(b-c)+c a(c-a)=k(a-b)$ $(b-c)(c-a)$.
Let $a=0, b=1, c=2$, we get
$$
0-2+0=k \cdot(-1)(-1)(2) .
$$

Therefore, $\mathbf{k}=-1$.
Thus, $a b(a-b)+b c(b-c)+c a(c-a)$ $=-(a-b)(b-c)(c-a)$.