160 The rectangular parallelepiped $A B C D-A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ has dimensions $a, b, c$ for its length, width, and height, respectively. The intersection of the two tetrahedra $A B^{\prime} C D^{\prime}$ and $A^{\prime} D C^{\prime} B$ is a polyhedron $V$. Then the volume of $V$ is $\qquad$
$160 \frac{a b c}{6}$. The tetrahedron $A B^{\prime} C D^{\prime}$ is a rectangular parallelepiped with 4 corners cut off, i.e., 4 tetrahedra: $B-A B^{\prime} C, D-A D^{\prime} C, C^{\prime}-C D^{\prime} B^{\prime}, A^{\prime}-B^{\prime} D^{\prime} A$. Each has a volume of $\frac{1}{6}$ of the rectangular parallelepiped.
So $V_{A B^{\prime} C^{\prime}}=\left(1-\frac{1}{6} \times 4\right) a b c=\frac{1}{3} a b c$.
The common part of the tetrahedra $A B^{\prime} C D^{\prime}$ and $A^{\prime} B C^{\prime} D$ is the tetrahedron $A B^{\prime} C D^{\prime}$ with four corners cut off, each of which is a tetrahedron, such as $A-O P Q$, where $O, P, Q$ are the midpoints of $B D, A^{\prime} B, A^{\prime} D$, which are also the midpoints of $A C, A B^{\prime}, A D^{\prime}$, so $V_{A}\left(P Q=\frac{1}{8} V_{A} \cdot C B^{\prime} D^{\prime}\right)$.
$$
V=\left(1-4 \times \frac{1}{8}\right) V_{A B^{\prime} C D^{\prime}}=\frac{1}{6} a b c \text {. }
$$
(Note: Let the centers of each face of the rectangular parallelepiped be $P, Q, R, S, T, O$, then the common part is an octahedron with vertices at $P, Q, R, S, T, O$. It is easy to see that its volume is $\frac{1}{3} \times\left(\frac{\sqrt{2}}{2}\right)^{2} a b c=\frac{1}{6} a b c$. )