More Hypercubes

Just as the volume of a cube is ${ (side\: length) }^{ 3 }$ , the volume of an n-dimensional hypercube is ${ (side\: length) }^{ n }$ . We know the side length of our cube is then $\sqrt [ n ]{ 81 }$ .

The farthest distance between any two vertices will be the distance between vertices on opposite corners of the cube (the line connecting them runs through its center). Arbitrarily setting one of the vertices to a location of $0$ on every axis makes the relative location of the other vertice $\sqrt [ n ]{ 81 }$ on every axis.

The n-dimensional distance formula is as follows for distances $d$ on each respective axis and $n$ dimensions: $\sqrt { { { d }_{ 1 } }^{ 2 }+{ { d}_{ 2 } }^{ 2 }+{ { d }_{ 3 } }^{ 2 }+...+{ { d }_{ n } }^{ 2 } }$ . Since the distance is the same in each dimension, and there are $n$ additions being made, we can construct a formula for the linear distance in terms of the dimension and volume. $\sqrt { n{ (\sqrt [ n ]{ volume } ) }^{ 2 } } =distance\\ \sqrt { n{ (\sqrt [ n ]{ 81 } ) }^{ 2 } } =2\sqrt { 6 } \\\sqrt[n]{81} \sqrt { n } =2\sqrt { 6 } \\ ...\\ n=\boxed { 8 }$

(I don't actually know a way to find $n$ from the simplified equation other than brief trial and error with different integers. Wolfram Alpha give $n$ in terms of a function I'm unfamiliar with. If anyone knows a simple way then please comment!)