Let $p(t) = t^3 + Ft^2 + Et + V$, where $F,E,V$ are the number of faces, edges, and vertices of a cube, respectively.
Factor $p(t)$ and explain your results in terms of generating functions.
A hint I got: First, you may wish to try factoring the corresponding polynomial for a square. That is, factor $t^2+Et+V$, where $E$ and $V$ are the number of edges (sides) and vertices (corners) of a square. Can you explain the result?
On
We are computing the product $(1+1+t)(1+1+t)(1+1+t)$, which we do by selecting one term from each factor in all $27$ possible ways, then summing the result.
Interpret the three factors as the $x$, $y$, and $z$ coordinates. We select the first $1$ if we want that coordinate to be 0, the other if we want it to be $1$. And we select the $t$ if we allow it to be anything.
With this construction, how many ways are there to get, e.g. an edge? We pick values for two of the coordinates, and allow the third to be anything. Every such combination will have exactly one $t$ in it, so this demonstrates that the number of edges equals the coefficient of $t$ in the generating function. The other coefficients are exactly analogous.
The point is that a cell of a cube is exactly determined by specifying coordinates in precisely this way: you decide which coordinates you want to restrict, and how.
The $(d+1)$-cube is the Cartesian product of the $d$-cube and the 1-cube aka the unit interval. Your obeservation is a reflection of the fact the "face generating function" of a Caresian product the product of the generating functions of its factors. So for the $d$-cube the generating function is $(t+2)^d$.
This is not a complete proof, because I have not verifed the "fact". This is reasonably straightforward when one of the factors is the unit interval, which is all that we need here.