The volume of a regular tetrahedron of side [math]a[/math] is [math]\frac{(\sqrt{2}a)^3}{24}[/math]. This construction helps you understand this formula. Twenty-four tetrahedra, all of the same volumes compose a cube. There are eight regular tetrahedra and sixteen non regular tetrahedra and each is actually a fourth of a regular octahedron. A step in this decomposition is the central octahedron, with the eight regular tetrahedra on each of its equilateral triangle faces, composing Kepler's [url=https://en.wikipedia.org/wiki/Stellated_octahedron]stella octangula[/url]. Then you add 4 non regular tetrahedra in each direction to compose the whole cube. Each tetrahedron has the same equilateral triangle basis and the same height (two applications of the Pythagorean theorem will show this), hence the same volume. Therefore, the volume of each of these 24 tetrahedra is the same as a 1/24th of the cube of side [math]\sqrt {2}a[/math].

Put the graphical window in full screen and play with the view, turn it, zoom in and control the animation.[br][br]Convince yourself that the tetrahedra all have the same volume. You can do it analytically or geometrically. A [url=https://www.youtube.com/playlist?list=PLFzaj-tjjVb96CBXctTgu_s9KCn7qJtTf]YouTube playlist[/url] with origami. [img]http://math.univ-lyon1.fr/irem/IMG/png/Cube24Tet.png[/img]