A cube has 8 vertices. Total ways to choose 4 vertices: $\binom{8}{4} = 70$. However, 6 sets of 4 vertices are coplanar (the 6 faces of the cube). Also, 6 more sets are coplanar (the 6 diagonal planes). So $70 - 6 - 6 = 58$ non-coplanar sets, each forming a tetrahedron.