2004
Problem - 3620
Let non-zero real numbers $a, b, c$ satisfy $a+b+c\ne 0$. If the following relations hold $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}$$
Find the value of $$\frac{(a+b)(b+c)(c+a)}{abc}$$
Suppose $$\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=k$$
Then $$k=\frac{(a+b-c)+(a-b+c)+(-a+b+c}{a+b+c}=1$$
Therefore $$\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=k+1=2$$
Finally, $$\frac{(a+b)(b+c)(c+a)}{abc}=2\times 2\times 2=\boxed{8}$$