1. Given that $a$, $b$, and $c$ are non-zero real numbers, satisfying
$$
\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c} \text {. }
$$

Then the value of $\frac{(a+b)(b+c)(c+a)}{a b c}$ is $\qquad$
2. -1 or 8.
Let $\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c}{c}=k$, then
$$
\begin{array}{l}
b+c=(k+1) a, c+a=(k+1) b, \\
a+b=(k+1) c .
\end{array}
$$

Thus, $2(a+b+c)=(k+1)(a+b+c)$.
Hence $a+b+c=0$ or $b+c=2 a, c+a=2 b, a+b$ $=2 c$.
Therefore, $\frac{(a+b)(b+c)(c+a)}{a b c}=-1$ or 8.