\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{b+c+a}{a}=\frac{c+a+b}{b}\)
Xét \(a+b+c\ne0\Rightarrow a=b=c\).Khi đó \(A=2\cdot2\cdot2=8\)
Xét \(a+b+c=0\)\(\Rightarrow\left\{\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Khi đó \(A=\frac{a+b}{a}\cdot\frac{a+c}{c}\cdot\frac{b+c}{b}=\frac{-c}{a}\cdot\frac{-b}{c}\cdot\frac{-a}{b}=-1\)