9.2. Let $x$, $y$, and $z$ be non-zero real numbers satisfying the equations: $\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}$. Find all possible values of the expression $\frac{(x+y)(y+z)(z+x)}{x y z}$. Justify that there can be no other values.
Solution: Let $S=x+y+z$. Then $\frac{x+y}{z}=\frac{S-z}{z}=\frac{S}{z}-1, \frac{y+z}{x}=\frac{S}{x}-1$, $\frac{z+x}{y}=\frac{S}{y}-1$, and the condition reduces to the equality: $\frac{S}{z}=\frac{\frac{S}{S}}{y}=\frac{S}{x}$. From this, either $x=y=z$, and then the desired expression equals 8, or $S=0$, and then it equals
$\left(\frac{S-x}{x}\right)\left(\frac{S-y}{y}\right)\left(\frac{S-z}{z}\right)=-1$. Both situations are possible: for example, the first one when $x=y=z=1$, the second one when $x=y=1, z=-2$.

Answer: 8 or -1.

Recommendations for checking:

| present in the work | points |
| :--- | :--- |
| Correct and justified answer | 7 points |
| Justified that only the values 8 and -1 are possible, but no examples are provided showing that both of these values are achieved | 6 points |
| Both possible values 8 and -1 are correctly found, but it is not shown that there are no others | 2 points |
| One of the possible values 8 or -1 is correctly found | 1 point |