+1253 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)
