#1**+2 **

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)

CoolStuffYT Feb 10, 2019