am i missing something? $\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}=\frac{(x+y)+(y+z)+(z+x)}{x+y+z}=2$ so $\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{x+y}{z}\cdot\frac{y+z}{x}\cdot\frac{z+x}{y}=\boxed{8}$ edit: see below, $\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}=-1$ also works, so $-1$ is another solution.

fruitmonster97 wrote: am i missing something? $\frac{x+y}{z}=\frac{y+z}{x}=\frac{z+x}{y}=\frac{(x+y)+(y+z)+(z+x)}{x+y+z}=2$ so $\frac{(x+y)(y+z)(z+x)}{xyz}=\frac{x+y}{z}\cdot\frac{y+z}{x}\cdot\frac{z+x}{y}=\boxed{8}$ What if $x + y+z =0$?

we cant divide for 0

Here is another solution: Observe that x^2+xy=z^2+yz and (x+y+z)(x-z)=0 1)x+y+z=0 2)x=z Do it for others and the all possible values are -1,8

We have $x^2+xy=z^2+zy,$ thus $(x-z)(x+z)+y(x-z)=0$ so $(x-z)(x+y+z)=0.$ In the case of $x=z,$ we have $\frac{2x}{y}=1+\frac{x}{y},$ thus it is clear that $\frac{x}{y}=1,$ and thus $\frac{x + y}{z}=\frac{y + z}{x}=\frac{z + x}{y}=2,$ thus the product is $8.$ Now when $x=-y+z,$ we have $\frac{y+z}{x}=-1,$ so in this case the product is $-1.$ So the values are $-1,$ or $8.$

The answer is $-1$ and $8$. Note that $$\frac{x+y}z=\frac{y+z}x \Rightarrow (x+y+z)(x-z)=0.$$The rest is easy.

parmenides51 wrote: Let $x, y, z$ be nonzero real numbers with $$\frac{x + y}{z}=\frac{y + z}{x}=\frac{z + x}{y}.$$Determine all possible values of $$\frac{(x + y)(y + z)(z + x)}{xyz}.$$ (Walther Janous) https://artofproblemsolving.com/community/c6h1610881p10068907 here