Assuming congruent edges, the regular octahedron has volume 4 times a regular tetrahedron. Any easy way to illustrate this is to consider two tetrahedra, a big one with edge length 2 and a small one with edge length 1. The big tetrahedron has volume 8 times the small one, because volume scales by the cube of the linear dimension.
Place 4 copies of the small one inside the big one, one inside each vertex. (Incidentally, this draws a Zelda Triforce on each face. The whole arrangement could be considered a 3-dimensional analogue of the Triforce, maybe a Tetraforce.)
The remaining space in the middle has the volume of 8-4 = 4 small tetrahedra. The remaining space in the middle is a regular octahedron with the same edge length as the small tetrahedra. Therefore, the octahedron has the volume of 4 tetrahedra.
Motivating question: how uneven in size are the components of the tetrahedral-octahedral honeycomb? Answer: they differ in volume by a ratio of 4, which is a lot compared the cubical honeycomb (or any parallelopiped) whose components are of course all the same size. However, such a volume ratio would correspond to a ratio of linear dimensions of cbrt 4 = 1.6, which might make it seem not so much. Incidentally, that cube root can be approximated by 16^3=2^12=4096.
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