This formula takes in the six edges of any tetrahedron. Think of four points at random in space, but not coplanar. Instead of calibrating this space in unit cubes, we have an "isotropic vector matrix" going, otherwise known as the CCP or FCC. Tetrahedrons and octahedrons fill space, of relative volume 1:4, and we consider a regular tetrahedron our unit of volume, frequenting a "duo-tet" cube (Kepler: defined by a Stella Octangula) of volume 3.
In this namespace, the tetrahedron reigns supreme as the topologically simplest of the Five Platonics, its own dual. That's why we wanted a way to compute tetravolumes easily and this formula does the job.
I should also mention that the REPL is checking the tetravolume of a specific shape, known as the S-module, here's what it looks like as a plane net.
Here's where it fits in the grand scheme of things (i.e. in the Concentric Hierarchy, with the Octahedron our volume 4):
That volume may be re-expressed in terms of Phi i.e. these two should be equal:
PHI = (1 + rt2(Decimal('5')))/2 print("smod.ivm_volume =", smodule()) print("smod check =", (1/2) * 1/(PHI ** 5))
There's a formula a lot like it in the history books, from the 1400s:
Heron's Formula for the Tetrahedron
Here's a picture to think with. Think of a, b, c coming from any apex, then the next three terms are a to b, b to c, c back to a. If the corner abc is of a regular tetrahedron (angle-wise), then the volume is abc in terms of tetravolumes, same as we'd get with unit cubes.
The abs( ) is not really necessary as the computation should fail entirely if the six edges do not describe a possible tetrahedron e.g. if you give a huge base and tiny edges that could never meet at an apex.