Unit Cube Diagonals

There are $12$ surface diagonals which each have a length of $\sqrt{2}$ thus the product of the surface diagonals is $(\sqrt{2})^{12}=2^6=64$ . There are $4$ space diagonals which each have a length of $\sqrt{3}$ thus the product of the surface diagonals is $(\sqrt{3})^4=3^2=9$ . Thus our answer is $64*9=\boxed{576}$