Đặt x=a-b; y=b-c; z=c-a
=>x+y+z=0
Ta có: \(\frac{(a-b)(a+b)}{b-c} = \frac{(b-c)(b+c)}{c-a} = \frac{(c-a)(c+a)}{a-b} = 1\)
=>\(\begin{cases}\dfrac{x(a+b)}{y}=1\\ \dfrac{y(b+c)}{z}=1\\ \dfrac{z(c+a)}{x}=1\end{cases}\)
=>a+b=y/x; b+c=z/y; c+z=x/z
\(P = \left(\frac{y}{x} + 1\right)\left(\frac{z}{y} + 1\right)\left(\frac{x}{z} + 1\right)\)
\(=\frac{x+y}{x}\cdot\frac{y+z}{y}\cdot\frac{z+x}{z}=\frac{-z}{x}\cdot\frac{-x}{y}\cdot\frac{-y}{z}=-1\)
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