If we consider "a" as the edge length, and "D" the cube's diagonal, we have that the square cube's diagonal is equal to the edge length's square plus the side diagonal (d) square (Pythagoras theorem)
a² + d² = D²
And since:
d² = a² + a²
Clearing a, we have:
a² = D²-d²
a² = D²-2a²
3a² = D²
a = √(D²/3)
Surface area is equal to 6·a², so the surface area will be 6·(D²/3) = 2D²
The volume is a³, so the volume will be √(D²/3)³ = √([tex] D^{6} [/tex]/3³) = D³/√27
