13.1.9 ** Let $P(e, \pi, \sqrt{5})$ be a point inside a cube whose vertices are the 8 points $(x, y, z)$ (where $x, y, z$ are 0 or 6). Find the volume of the intersection of the polyhedron with the original cube, where the polyhedron's vertices are the 6 symmetric points of $P$ with respect to the cube's faces.
Let point $P(a, b, c)$ be a point inside a cube. The points symmetric to $P$ with respect to each face form an octahedron. One of its faces, for example, the triangle formed by $A(-a, b, c), B(a,-b, c), C(a, b,-c)$, intersects the original cube at $(0,0, c),(a, 0,0),(0, b, 0)$. Therefore, the volume of the tetrahedron cut off is $\frac{1}{6} a b c$. Similarly, the volume of the tetrahedra cut off by other faces can be obtained, and the sum of the cut-off volumes is $V=\frac{1}{6} \sum x_{i} y_{j} z_{k}$, where $i, j, k$ are 1 or 2. Since $x_{1}+x_{2}=y_{1}+y_{2}=z_{1}+z_{2}=6$, we have $V=\frac{1}{6}\left(x_{1}+x_{2}\right)\left(y_{1}+y_{2}\right)\left(z_{1}+z_{2}\right)=6^{2}$. Therefore, the volume of the common part is $6^{3}-6^{2}=180$.