TITLE: Hole inside cube with tetrahedrons at corners?
QUESTION [14 upvotes]: Given is a unit cube with a tetrahedron at each corner, as shown here for one corner out of the $8$ :

It is noticed that the tetrahedrons are not disjoint. Because I cannot look through the cube, I have great difficulty imagining whether there is a hole left inside or not. If there is a hole, what then is the shape of that hole? And what then is the volume of that hole?
The volume of one tetrahedron is $1/6$ . This would make a total of $\,8/6\,$ if they were disjoint, but - as I've said - they are not. Apart from the facts some sort of proof would be nice.

REPLY [2 votes]: Split your cube:

1 octahedron in the middle



8 regular tetrahedra on each face of the octahedron to obtain a stellated octahedron:



12 tetrahedra (green) between adjacent pairs of regular tetrahedra (blue) to obtain a cube:


Split your tetrahedra:

Notice that the tetrahedron you describe is the union of one blue tetrahedron and 3 green tetrahedra.
Each blue tetrahedron is used once.
Each green tetrahedron is used twice.