44. Prove that from the eight vertices of a cube, four vertices can be chosen in two ways such that they form the vertices of a regular tetrahedron. Calculate the ratio of the volume of such a tetrahedron to the volume of the cube. Also, calculate the volume of the figure that is the intersection of these two tetrahedrons, and the volume of the figure that is their union. (Assume that the length of the edge of the cube is \( a \)).
44. In the cube $A B C D A^{\prime} B^{\prime} C^{\prime} D^{\prime}$, the face diagonals $[A C],\left[C D^{\prime}\right],\left[D^{\prime} A\right]$, $\left[A B^{\prime}\right],\left[B^{\prime} C\right],\left[B^{\prime} D^{\prime}\right]$ serve as the edges of a regular tetrahedron, while the other 6 face diagonals serve as the edges of another regular tetrahedron. The volume of one of these tetrahedrons can be obtained by subtracting from the volume of the cube the volumes of four pyramids, where the base is half the face of the cube and the height is the edge of the cube. The volume of such a pyramid is $\frac{1}{3} a \cdot \frac{1}{2} a^{2}=\frac{1}{6} a^{3}$, so the volume of the tetrahedron is: $a^{3}-4 \cdot \frac{1}{6} a^{3}=\frac{1}{3} a^{3}$.

The intersection of the considered tetrahedrons is an octahedron, whose vertices are the centers of the faces of the cube. This octahedron can be obtained by cutting off four tetrahedrons with edges half the length from one of the larger tetrahedrons. The volume of each of these four tetrahedrons is 8 times smaller than the volume of the larger tetrahedron, and thus the volume of the remaining octahedron is half the volume of the tetrahedron, i.e., $\frac{1}{6} a^{3}$.

The volume of the polyhedron, which is the union of the tetrahedrons, can be obtained by attaching four tetrahedrons with half the edge length to one of the tetrahedrons (one to each face); with this addition, the volume of the tetrahedron increases by one and a half times, i.e., it becomes equal to: $\frac{3}{2} \cdot \frac{1}{3} a^{3}=\frac{1}{2} a^{3}$.