AUT_ABooklet_2023

Answer. The only possible values are $8$ and $-1$.

We add $1$ to the equations and obtain $$\frac{x+y+z}{z}=\frac{x+y+z}{x}=\frac{x+y+z}{y}.$$ For $x+y+z=0$, this is clearly true, and the expression in the problem statement becomes $$\frac{(-z)(-x)(-y)}{xyz}=-1.$$ For $x+y+z\ne 0$, we get $$\frac{1}{z}=\frac{1}{x}=\frac{1}{y}.$$ and therefore $x=y=z$. In this case, our expression becomes $8$. The values $-1$ and $8$ are attained because any triple with $x+y+z=0$ resp. $x=y=z$ that does not contain a zero works. (Theresia Eisenkölbl) ☐