Example 15. Factorize:
$$
(a-b)^{3}+(b-c)^{3}+(c-a)^{3} .
$$
Solution 1 Using term splitting and substitution
$$
\begin{array}{l} 
\because a-b=-(b-c)-(c-a), \\
\therefore \text { original expression }=-[(b-c)+(c-a)]^{3} \\
\quad+(b-c)^{3}+(c-a)^{3} \\
=-3(b-c)^{2}(c-a)-3(b-c) \\
\cdot(c-a)^{2} \\
= 3(a-b)(b-c)(c-a) .
\end{array}
$$

Solution 2 Using factor theorem.
$\because$ When $a=b$, the original expression $=0$,
$\therefore$ this polynomial has a factor $a-b$.
$\because$ This polynomial is a cyclic symmetric polynomial in $a, b, c$,
$\therefore$ it also has factors $b-c, c-a$.
Let the original expression $\equiv k(a-b)(b-c)(c-a)$.
Comparing the coefficients of $a^{2} b$ on both sides, we get $k=3$.
$\therefore$ The original expression $=3(a-b)(b-c)(c-a)$.