Giải:

$-$ Ta có:

$\displaystyle \frac{a+b-c}{c} = \frac{b+c-a}{a} = \frac{c+a-b}{b}$

$\displaystyle \iff \frac{a+b-c}{c} + 2 = \frac{b+c-a}{a} + 2 = \frac{c+a-b}{b} + 2$

$\displaystyle \iff \frac{a+b+c}{c} = \frac{a+b+c}{a} = \frac{a+b+c}{b}$

$-$ TH1:

$a + b + c = 0 \implies \begin{cases} a + b = -c \\ b + c = -a \\ c + a = -b \end{cases}$

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

$\displaystyle \implies A = \frac{(-c)(-a)(-b)}{abc} = \frac{-abc}{abc} = -1$

$-$ TH2: $a + b + c \neq 0$

$\displaystyle \implies a = b = c$

$\displaystyle \implies A = \left(1 + \frac{a}{a}\right)\left(1 + \frac{a}{a}\right)\left(1 + \frac{a}{a}\right) = (1 + 1)(1 + 1)(1 + 1) = 2 \cdot 2 \cdot 2 = 8$

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