Five. (Full marks 10 points) Given $\frac{a+b-c}{c}=\frac{a-b+c}{b}=$ $=\frac{a+b+c}{a}$. Find the value of $\frac{(a+b)(b+c)(c+a)}{a b c}$.

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Let $\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$.
Then
\[
\begin{array}{l}
a+b=(k+1) c, \\
b+c=(k+1) a, \\
c+a=(k+1) b . \\
(1)+(2)+(3) \text{ gives } \\
(k-1)(a+b+c)=0 .
\end{array}
\]

When $k=1$, the original expression $=8$;
When $k \neq 1$, $a+b+c=0$, the original expression $=-1$.