\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=2\)
\(\Rightarrow a+b-c=2c\Rightarrow a+b=3c\)
\(b+c-a=2a\Rightarrow b+c=3a\)
\(c+a-b=2b\Rightarrow c+a=3b\)
Ta có: \(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(a+\frac{c}{b}\right)\)
\(=\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}\)
\(=\frac{3c}{a}.\frac{3b}{c}.\frac{3a}{b}\)
\(=27\)