I'll be eternally grateful if you can answer all these geometry questions!

For the challenge...

The side length of the cube is $66^{1/3}=\sqrt[3]{66}$

The volume of a square pyramid is $a^2\frac{h}{3}$, where $a$ is the base edge and $h$ is the height.

Of course, they are all $\sqrt[3]{66}$.

So...

$\sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22$ in^3 is the volume.

You are very welcome!

:P

(that was a lot of LaTeX for me: \sqrt[3]{66}^2\times \frac{\sqrt[3]{66}}{3}=66^{\frac{2}{3}}\times \frac{66^{\frac{1}{3}}}{3}=\frac{66}{3}=22 was what i entered)

Do you have a diagram for the previous problems...