Đặt \(\frac{a+b}{c}=\frac{a+c}{b}=\frac{b+c}{a}=k\Rightarrow\left\{{}\begin{matrix}a+b=ck\\a+c=bk\\b+c=ak\end{matrix}\right.\)
Có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(\Rightarrow M=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}\\ \Rightarrow M=\frac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{b\cdot c\cdot a}\\ M=\frac{ck\cdot bk\cdot ak}{b\cdot c\cdot a}=\frac{a\cdot b\cdot c\cdot k^3}{a\cdot b\cdot c}=k^3\)
Vậy M = k3