- 1. $\{3\}$.

Assume $x>y$. Then
$$
\begin{array}{l}
x^{3}>y^{3} \Rightarrow \frac{1}{y+2016}z \\
\Rightarrow y^{3}>z^{3} \Rightarrow \frac{1}{z+2016}x,
\end{array}
$$

which is a contradiction.
Assume $x<y$.
Similarly, $y<z, z<x$, which is also a contradiction.
Therefore, $x=y$. Consequently, $y=z$.

Hence $x=y=z \Rightarrow \frac{x^{3}+y^{3}+z^{3}}{x y z}=3$.
1. Let $x, y, z > 0$, and
$$
\begin{array}{l}
x^{3}+\frac{1}{y+2016}=y^{3}+\frac{1}{z+2016} \\
=z^{3}+\frac{1}{x+2016} .
\end{array}
$$

Then the set of values for $\frac{x^{3}+y^{3}+z^{3}}{x y z}$ is