5. For natural numbers $m$ and $n$, where $m<n$, it holds that

$$
\left(\frac{m}{n}\right)^{3}=\overline{0, x y z x y z x y z \ldots}
$$

where $x, y$, and $z$ are some digits (not necessarily distinct), and the block $\overline{x y z}$ repeats periodically infinitely many times. Determine all possible values for $\frac{m}{n}$.
5. Note:

$$
1000\left(\frac{m}{n}\right)^{3}=\overline{x y z, x y z x y z x y z \cdots}=\overline{x y z}+\overline{0, x y z x y z x y z \cdots}=\overline{x y z}+\left(\frac{m}{n}\right)^{3}
$$

then we get $999\left(\frac{m}{n}\right)^{3}=\overline{x y z}$, i.e., $999 m^{3}=\overline{x y z} \cdot n^{3}$. Clearly, we can assume that $m$ and $n$ are coprime. From this, it follows that $n^{3} \mid 999=3^{3} \cdot 37$. Since $n>1$ (because $m<n$), the only possibility is $n=3$. Then $m$ can be $1$ or $2$. In the first case, we get $\overline{x y z}=\frac{999}{27}=37=\overline{037}$, and in the second case, $\overline{x y z}=\frac{999 \cdot 8}{27}=296$. Therefore, both possibilities for $m$ indeed provide solutions to the problem, so $\frac{m}{n}$ can be $\frac{1}{3}$ or $\frac{2}{3}$.