Since this question remained without answers even after a bounty, I thought it might be time to ask it here.

For which pyramid can you compute the volume from simple cut-and-glue processes? The Dehn invariant naturally gives the answer, but I failed to turn this in an algorithm. Here are the pyramids, I know of, whose volume is computable by elementary operations:

• take a cube and divide into six pyramids from its center (or three pyramids given the 120° symmetry along a diagonal). These pyramids can be subdivided and glued into further pyramids, but there are not so many possibilities.

• take a trigonal trapezohedron whose faces are all rhombic. This is not a right prism, but oblique prisms can also be cut-and-glued to compute their volume. Because of his symmetry group you can cut it into three pyramidal pieces (oblique pyramids). Since the small angle of the rhombus can be $\in ]0, \pi/2[$ (at $\pi/2$ it's just a cube), this gives an infinite family of pyramids. (These oblique pyramids have a symmetry: you can cut them further in half.)

But the that's all I could find. Are there any other? and if so, what is the cut-and-glue process?

Some background:

• The volume of pyramids is discussed here in Proposition 3 to 5 of Euclid's book

• The fact there was no "simple" proof irritated Gauß and later Hilbert. It was purportedly the inspiration for his 3rd Problem, which was solved by Dehn (using his invariant)

• There is an earlier (and largely ignored) solution of Hilbert's 3rd problem by a certain Birkenmayer, see this paper.

• In two dimensions the Wallace–Bolyai–Gerwien theorem answers the question completely. Beside the Dehn invariant (which shows this is not the case in three dimensions), the Banach-Tarski paradox is also a reminder that volume in three dimensions are tricky.

[EDIT: To clarify simple cute-and-glue (as asked in the comments). One starts with the class $(C)$ of solids (at the beginning it only includes rectangular prisms). You can add another polyhedron $P$ to $(C)$ if:

a number $k \neq 0$ of copies of $P$ can be obtained by cutting solids $\{S_i\}_{i=1,\ldots,s}$ from $(C)$ [possibly many copies of the same solid] into finitely many polyhedral pieces and reassembling these pieces together into the $k$ copies $P$ and a possibly empty collection of solids $\{T_i\}$ with $T_i$ belonging to $(C)$. In that case, $\mathrm{vol}(P) = \tfrac{1}{k} \big( \sum \mathrm{vol}(S_i) - \sum \mathrm{vol}(T_i) \big)$, and so $P$ can be added to $(C)$.

Reassembling means apply isometries of $\mathbb{R}^3$ (rotations, reflections and translations); two solids are congruent if one is the image of the other under such transformations and a copy is also the image of a solid under such transformations. This definition is closely related to scissors-congruence.