Giải thích các bước giải:
Ta có :
$\dfrac{a+b-c}{3c}=\dfrac{b+c-a}{3a}=\dfrac{c+a-b}{3b}$
$\to\dfrac{a+b}{3c}-\dfrac13=\dfrac{b+c}{3a}-\dfrac13=\dfrac{c+a}{3b}-\dfrac13$
$\to\dfrac{a+b}{3c}=\dfrac{b+c}{3a}=\dfrac{c+a}{3b}$
$\to\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+c+a}{a+b+c}=\dfrac{2(a+b+c)}{a+b+c}=2$
$\to \dfrac{a+b}{c}.\dfrac{b+c}{a}.\dfrac{c+a}{b}=2^3$
$\to \dfrac{(a+b)(b+c)(c+a)}{abc}=8$
$\to \dfrac{a+b}{a}.\dfrac{b+c}{b}.\dfrac{c+a}{c}=8$
$\to (1+\dfrac ba)(1+\dfrac cb)(1+\dfrac ac)=8$
$\to B=8$