We rotate a square around one of its axes of symmetry. The surface area of the resulting solid of revolution is $F$, and its volume is $V$. What values can the ratio $F^{3} / V^{2}$ take?
A square has two types of axes of symmetry, one is the line bisecting the side, and the other is the diagonal.

Let the side of the square be $2a$. If we rotate the square around one of its side bisectors, a cylinder is formed with a height of $2a$ and a radius of the base circle of $a$.

The surface area is $F=2a^{2}\pi+2a\pi \cdot 2a=6a^{2}\pi$, and the volume is $V=a^{2}\pi \cdot 2a=2a^{3}\pi$. The desired ratio is

$$
\frac{F^{3}}{V^{2}}=\frac{\left(6a^{2}\pi\right)^{3}}{\left(2a^{3}\pi\right)^{2}}=54\pi
$$

In the second case, when the square is rotated around one of its diagonals, we get two congruent cones. The radius and height of the base circle of the cone are $a\sqrt{2}$. The surface area, $F$, is the sum of the lateral surface areas of the two cones.

$$
F=\frac{2a\sqrt{2}\pi \cdot 2a}{2} \cdot 2=4a^{2}\sqrt{2}\pi
$$

The volume is

$$
V=2 \cdot \frac{(a\sqrt{2})^{2}\pi \cdot a\sqrt{2}}{3}=\frac{4a^{3}\pi\sqrt{2}}{3}
$$

Thus,

$$
\frac{F^{3}}{V^{2}}=\frac{2 \cdot 64a^{6}\pi^{3}\sqrt{2}}{\frac{2 \cdot 16a^{6}\pi^{2}}{9}}=36\sqrt{2}\pi
$$

Therefore, the ratio $\frac{F^{3}}{V^{2}}$ can take two values: $36\sqrt{2}\pi$ and $54\pi$.