Width of layer formed from sphere $r = n$ fitted inside sphere $r = 4n$

Question

If a volume equivalent to a solid sphere of radius $n$ is pushed onto the surface of a sphere with radius $4n$, how wide is the layer that forms, expressed relative to the radius $n$?

Answer 1

Consider a new radius of $r$ for the amount inside the sphere that doesn't include the volume which has been pushed onto the inside of the sphere of radius $4n$. Its volume will be the original sphere volume less the volume of the matter added to its insides. Thus, using the formula for the volume of a sphere, we get that

$$\cfrac{4}{3}\pi r^3 = \cfrac{4}{3}\pi \left(4n\right)^3 - \cfrac{4}{3}\pi n^3 = \cfrac{4}{3}\pi \left(63n^3\right) \tag{1}\label{eq1}$$

Dividing both sides by $\frac{4}{3}\pi$ and taking cube roots, we get that

$$r = \sqrt[3]{63} n \tag{2}\label{eq2}$$

Thus, the width of the layer that forms would be the difference between the original & new radii, i.e.,

$$4n - r = \left(4 - \sqrt[3]{63}\right)n \tag{3}\label{eq3}$$