It is well known that an ordinary cube can be made of 5 tetrahedra: *1 regular tetrahedron* at the center, plus *4 trirectangular tetrahedra* (*trt*) glued to the 4 faces of the former.
Let the edge of the cube be *2*. Then the edges of the *regular tetrahedron* (in the interior) will have length = 2.sqrt(2) (the diagonal of a face of the cube) & its volume will be 8/3. Now, the *trt* has the same base as a face of the reg. tetrahedron & 1/2 its altitude, so its volume is 4/3. All put together:
8/3 (*reg. tetrah.*)+ 4. 4/3 (*trt*s)= 24/3 = 8 (volume of the *cube*)