Ta có \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

=> \(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

Nếu a + b + c = 0

=> a + b = -c

b + c = -a

a + c = -b

Khi đó P = \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{a+c}{c}=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)

Nếu a + b + c \(\ne\)0

=> \(\frac{1}{a}=\frac{1}{b}=\frac{1}{c}\)

=> a = b = c

Khi đó P \(\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=2.2.2=8\)

Vậy khi a + b + c = 0 thì P = -1

khi a + b + c  \(\ne\)0 thì P = 8

Theo đề ra\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)

\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

Mà: a + b + c khác 0  => a = b = c

=> P = (1 + 1)(1 + 1)(1 + 1) = 2 . 2 . 2 = 8

a + b + c = 0 => c = -a - b ; b= -a - c ; a =  - b - c 

Thay vào Q ta có :

\(Q=\frac{1}{a^2+b^2-\left(a+b\right)^2}+\frac{1}{b^2+c^2-\left(b+c\right)^2}+\frac{1}{a^2+c^2-\left(a+c\right)^2}\)

\(Q=\frac{1}{a^2+b^2-a^2-b^2-2ab}+\frac{1}{b^2+c^2-b^2-c^2-2bc}+\frac{1}{c^2+a^2-c^2-a^2-2ac}\)

\(Q=\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}=\frac{c+a+b}{-2abc}=0\)