15.26. Into how many parts do the planes of the faces divide the space a) of a cube; b) of a tetrahedron?
15.26. The planes of the faces of both polyhedra intersect only along the lines containing their edges. Therefore, the space divided into parts, into which the space is split, has a common point with the polyhedron. Moreover, to each vertex, each edge, and each face, one can correspond exactly one adjacent part to it, and these will exhaust all parts, except for the interior of the polyhedron. Thus, the required number is $1+\mathrm{B}+$ $+\Gamma+P$. For a cube, it is $1+8+6+12=27$, and for a tetrahedron, it is $1+4+4+6=15$.