A sphere tangent to the edges of a $4 \mathrm{~cm}$ cube is inscribed. One vertex of the cube is $A$. What is the volume of the region consisting of points that are inside the cube, outside the sphere, and closest to $A$ among the vertices of the cube?
Solution. Considering the entire cube, the volume of the part consisting of points inside the cube but outside the sphere is

$$
a^{3}-\frac{4}{3} \pi\left(\frac{a}{2}\right)^{3}
$$

where $a=4 \mathrm{~cm}$, the edge length of the cube.

If we impose the constraint that the points are closest to the vertex $A$ of the cube, then the volume of the part consisting of these points will be one-eighth of the previous quantity. The eight vertices of the cube are indeed equivalent, and the points that are equidistant from two or more vertices together occupy only a zero volume within the cube.

The sought volume is therefore

$$
\frac{1}{8} a^{3}-\frac{1}{48} \pi a^{3}=8-\frac{4}{3} \pi \approx 3.8112 \mathrm{~cm}^{3}
$$