Đáp án:
$x^2-y=y^2-z$
$→x^2-y^2=y-z$
$→(x-y)(x+y)=y-z$
$→$ $x+y=\dfrac{y-z}{x-y}$
$→$ $x+y+1=\dfrac{y-z+x-y}{x-y}=$ $\dfrac{x-z}{x-y}$
Tương tự: $y^2-z=z^2-x$
$→y+z+1=$ $\dfrac{y-x}{y-z}$
$x^2-y=z^2-x$
$→x+z+1=$ $\dfrac{y-z}{x-z}$
$→$ $M = ( x+y+1)(y+z+1)(z+x+1)=$ $\dfrac{x-z}{x-y}.$$\dfrac{y-x}{y-z}.$ $\dfrac{y-z}{x-z}=-1$
Vậy $M=-1$