1. Factorize $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}$
Let $x=b-c, y=c-a, z=a-b$.
$$
\begin{array}{l}
\text { By } x^{3}+y^{3}+z^{3}-3 x y z \\
=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right) \text {. }
\end{array}
$$

When $x+y+z=0$, $x^{3}+y^{3}+z^{3}=3 x y z$.
Therefore, the original expression $=3(a-b)(b-c)(c-a)$.