Bài giải:

Ta có: $\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}$ = $\frac{a+b-c+b+c-a+c+a-b}{a+b+c}$ = $\frac{a+b+c}{a+b+c}$ = 1

Suy ra: $\frac{a+b-c}{c}$ = 1

$\Rightarrow $ a + b – c = c

a + b = 2c

$\Rightarrow $ $\frac{a+b}{c}=2$

Chứng minh tương tự: $\frac{c+a}{b}$ = 2; $\frac{b+c}{a}$ = 2

B = $\left( 1+\frac{b}{a} \right)\left( 1+\frac{a}{c} \right)\left( 1+\frac{c}{b} \right)$

B = $\frac{a+b}{a}.\frac{c+a}{c}.\frac{b+c}{b}$

B = $\frac{a+b}{c}.\frac{c+a}{b}.\frac{b+c}{a}$ = 2. 2. 2 = 8

Vậy đáp án đúng là: C.