835. A three-digit number is equal to the cube of its unit digit. Find all such numbers.
Let's denote a three-digit number as $\overline{x y z}$. Then

$$
\overline{x y z}=z^{3}, \quad 10 \cdot \overline{x y}+z=z^{3}
$$

In which cases does the cube of the digit $z$ end in $z$? This is possible when $z=0,1,4,5$, 6, and 9. However, the values $z=0, z=1$, and $z=4$ are obviously not valid.

If $z=5$, then

$$
10 \cdot \overline{x y}+5=125, \quad \overline{x y}=12
$$

We have found the number 125.

If $z=6$, then

$$
10 \cdot \overline{x y}+6=216, \quad \overline{x y}=21
$$

We have the number 216.

If $z=9$, then

$$
10 \cdot \overline{x y}+9=729, \quad \overline{x y}=72
$$

We have found another number: 729.

Answer: $125,216,729$.