While your volume formula is correct, I don't see a way to cut a general parallepiped into 6 identical tetrahdtra. In fact, I don't see how it can be accomplished with a cube.
However these 6 tets are not necessarily congruent, when I commented I was visualizing the case when the parallelepiped has equal side lengths, but in the general case the 6 tetrahedra are not congruent. However each of the 6 is congruent to at least 1 other out of the 6, since the parallepiped has reflection symmetry.
Start with the cube, and notice that when you deform it into the general parallelepiped, the volumes of all the tetrahedra stay equal to eachother.