Which are the perfect cubes that divide $9^{4}$?
A perfect cube is a number of the form $a^{3}$, where $a$ is a natural number. Since $9^{4}=\left(3^{2}\right)^{4}=3^{8}$, the perfect cubes that divide $3^{8}$ are: $3^{3}$ and $\left(3^{2}\right)^{3}=3^{6}$.