Giải thích các bước giải:

Ta có:

$\dfrac{a-2b+c}b=\dfrac{b-5c+a}{2c} =dfrac{c-8a+b}{3a}$

$\to \dfrac{a+b+c-3b}b=\dfrac{b+c+a-6c}{2c} =dfrac{c+a+b-9a}{3a}$

$\to \dfrac{a+b+c}b-3=\dfrac{a+b+c}{2c}-3=\dfrac{a+b+c}{3a}-3$

$\to \dfrac{a+b+c}b=\dfrac{a+b+c}{2c}=\dfrac{a+b+c}{3a}$

Nếu $a+b+c=0$

$\to a+b=-c, b+c=-a, c+a=-b$

Ta có:

$P=(1+\dfrac ab)(1+\dfrac bc)(1+\dfrac ca)$

$\to P=\dfrac{a+b}b.\dfrac{b+c}c.\dfrac{c+a}a$

$\to P=\dfrac{-c}b.\dfrac{-a}c.\dfrac{-b}a$

$\to P=-1$

Nếu $a+b+c\ne 0$

$\to b=2c=3a$

$\to \dfrac{a}b=\dfrac13$

      $\dfrac{b}c=2$

      $\dfrac{c}a=\dfrac32$

$\to P=(1+\dfrac13)(1+2)(1+\dfrac32)=10$