Ta có \(\frac{\left(a+b\right)\left(a-b\right)}{b-c}=1\Leftrightarrow a+b=\frac{b-c}{a-b}\).
Tương tự: \(b+c=\frac{c-a}{b-c};c+a=\frac{a-b}{c-a}\).
Do đó: \(P=\left(\frac{b-c}{a-b}+1\right)\left(\frac{c-a}{b-c}+1\right)\left(\frac{a-b}{c-a}+1\right)=\frac{a-c}{a-b}.\frac{b-a}{b-c}.\frac{c-b}{c-a}=-1\).