This year's gift idea from BabyMath consists of a series of nine colored plastic containers of decreasing size, alternating in shape like a cube and a sphere. All containers can open and close with a convenient hinge, and each container can hold just about anything next in line. The largest and smallest container are both cubes. Determine the relationship between the edge lengths of these cubes.
1. **Understanding the problem**: We have a series of nine containers, alternating between cubes and spheres, with the largest and smallest containers being cubes. We need to determine the relationship between the edge lengths of the largest and smallest cubes.

2. **Setting up the relationship**: Let's denote the edge length of the largest cube as \( s_1 \). The next container is a sphere inscribed in this cube. The diameter of this sphere is equal to the edge length of the cube, \( s_1 \).

3. **Inscribing a cube in a sphere**: The next container is a cube inscribed in this sphere. The edge length of this cube, \( s_2 \), can be found by noting that the diagonal of the cube is equal to the diameter of the sphere. For a cube with edge length \( s_2 \), the space diagonal is \( \sqrt{3}s_2 \). Therefore, we have:
   \[
   \sqrt{3}s_2 = s_1 \implies s_2 = \frac{s_1}{\sqrt{3}}
   \]

4. **Continuing the pattern**: The next container is a sphere inscribed in the cube with edge length \( s_2 \). The diameter of this sphere is \( s_2 \). The next cube, with edge length \( s_3 \), is inscribed in this sphere. Thus:
   \[
   \sqrt{3}s_3 = s_2 \implies s_3 = \frac{s_2}{\sqrt{3}} = \frac{s_1}{(\sqrt{3})^2} = \frac{s_1}{3}
   \]

5. **Generalizing the pattern**: We can see that each subsequent cube's edge length is divided by \( \sqrt{3} \) compared to the previous cube. Therefore, for the \( n \)-th cube, the edge length \( s_n \) is given by:
   \[
   s_n = \frac{s_1}{(\sqrt{3})^{n-1}}
   \]

6. **Finding the edge length of the 5th cube**: Since the sequence alternates between cubes and spheres, the 5th cube is the 9th container. Thus, we need to find \( s_5 \):
   \[
   s_5 = \frac{s_1}{(\sqrt{3})^{5-1}} = \frac{s_1}{(\sqrt{3})^4} = \frac{s_1}{3^2} = \frac{s_1}{9}
   \]

Conclusion:
\[
\boxed{\frac{s_1}{9}}
\]