You should find at least 3 different decompositions. can you do it in more? why? It turns out that there are, up to symmetry, six possible ways of decomposing a regular cube into tetrahedra.

Just as in 3-dimensional space tetrahedra are the building blocks of our puzzles, one can imagine that a similar building block must exist for 4 or more dimensions. How would the analogue of a tetrahedron in dimension 4 look like? It must have 5 vertices (one more than the dimension of the ambience space...) and all of its vertices must be connected by edges. Another property is that every subset of 4 vertices must define a facet and so on. A simplex of dimension d is the generalization for d-dimensional Euclidean space of the 3-dimensional tetrahedron. One can again consider the question of decomposing polytopes in higher dimensions into simplices. In particular the decompositions of higher dimensional cubes into smallest number of simplices is a notoriously hard problem. So far we only know the answer of this question up to seven dimensions.

The general question of counting the different ways of decomposing a convex polyhedra is very hard and we only know the answer in a few cases. You may want to think of the analogue problem in the plane for simplicity: how many ways are there to decompose a n-gon into triangles? For example, what happens in the case of a hexagon? A simple explanation of how to recover the formula for the number of triangulations of a convex n-gon can be found in the book Introductory Combinatorics by Richard Brualdi (2nd ed. New York, North-Holland, 1992). A delightful explanation (in a delightful book!) about the deep mathematical structure behind the decompositions of polyhedra can be found in the last chapter of the book by Guenter Ziegler, Lectures on Polytopes (New York : Springer-Verlag, 1995).