Xét $x+y+z=0$
$\to \begin{cases}x+y=-z\\x+z=-y\\y+z=-x\end{cases}$
Thay vào $B$, ta có:
$B=(1+\dfrac{x}{y})(1+\dfrac{y}{z})(1+\dfrac{z}{x})$
$=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{x+z}{x}$
$=\dfrac{-z}{y}\times\dfrac{-x}{z}\times\dfrac{-y}{x}$
$=-1$
Xét $x+y+z\ne0$
Ta có:
$\dfrac{y+z-x}{x}=\dfrac{x+z-y}{y}=\dfrac{x+y-z}{z}$
$\to \dfrac{y+z}{x}-1=\dfrac{x+z}{y}-1=\dfrac{x+y}{z}-1$
$\to \dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{x+y}{z}$
Theo tính chất dãy tỉ số bằng nhau, ta có:
$\dfrac{y+z}{x}=\dfrac{x+z}{y}=\dfrac{x+y}{z}=\dfrac{y+z+x+z+x+y}{x+y+z}=2$
$\to \begin{cases}\dfrac{y+z}{x}=2\\\dfrac{x+z}{y}=2\\\dfrac{x+y}{z}=2\end{cases}$
$\to \begin{cases}y+z=2x\\x+z=2y\\x+y=2z\end{cases}$
Thay vào $B$ ta có:
$B=(1+\dfrac{x}{y})(1+\dfrac{y}{z})(1+\dfrac{z}{x})$
$=\dfrac{x+y}{y}\times\dfrac{y+z}{z}\times\dfrac{x+z}{x}$
$=\dfrac{2z}{y}\times\dfrac{2x}{z}\times\dfrac{2y}{x}$
$=2\times2\times2$
$=8$
