The bases of the parallelepiped are squares with side $b$, and all lateral faces are rhombuses. One of the vertices of the upper base is equidistant from all vertices of the lower base. Find the volume of the parallelepiped.
Let the vertex $B 1$ of the upper base $A 1 B 1 C 1 D 1$ of the parallelepiped $A B C D A 1 B 1 C 1 D 1$ be equidistant from all vertices of the lower base $A B C D$. Then the orthogonal projection of the vertex $B 1$ onto the lower base is the center $O$ of the square $A B C D$, and the quadrilateral pyramid $A B C D B 1$ with vertex $B 1$ is a regular pyramid. From the right triangle OBB1 we find that

$$
B 1 O=\sqrt{B B_{1}^{2}-O B^{2}}=\sqrt{b^{2}-\frac{b^{2}}{2}}=\frac{b}{\sqrt{2}}
$$

and since $B 1 O$ is the height of the parallelepiped, then

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot B 1 O=b 2 \cdot \frac{b}{\sqrt{2}}=\frac{b^{3}}{\sqrt{2}}
$$

## Answer