ta có \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)

\(\Rightarrow\dfrac{a+b}{c}-1=\dfrac{b+c}{a}-1=\dfrac{c+a}{b}-1\)

\(\Rightarrow\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)

+) với a + b + c = 0

\(\Rightarrow\left\{{}\begin{matrix}a=-\left(b+c\right)\\b=-\left(a+c\right)\\c=-\left(a+b\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)

+) với a + b + c ≠ 0

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\c+a=2b\end{matrix}\right.\)

\(\Rightarrow\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)