On the cubic complexity of three-dimensional polyhedra

A cubulation of a three-dimensional polyhedron $P$ is understood as a finite family of copies of the standard oriented cube in $\mathbb R^3$ and of orientation-changing isometries of its faces such that the result of gluing together these isometries of the cubes is homeomorphic to $P$. We prove that any three-dimensional polyhedron represented by a cubulation consisting of $n$ cubes possesses a standard triangulation consisting of $6n$ tetrahedra.