I am searching for an equation to extend a arbitrary multidimensional cuboid in such a way it stays a cuboid. Using equations like e.g. : $f + b^2 = c^2$, $f + b^3 = c^3$, ... $f + b^n = c^n$. I am looking for $f$ for which applies $f + b^n = c^n$ with $f,b,c,n \in N$ and $b>1$, $c>1$.
The figures below show examples. In the 2-dimensional example the 3x3 square can be arranged along the 4x4 square which results in a 5x5 square. In the 3-dimensional cuboid shown as green slices, the additional elements (grey) can be alligned around each slice like in the 2-dimensional square. But also additional slices must be added.
The $x$ is the amount of layers of grey elements added to the square in the 2-dimensional example.
In the 1-dimensional case: $f=(b+x)^1-b^1 = x^1$
In the 2-dimensional case: $f=(b+x)^2-b^2=x2b+x^2$
In the 3-dimensional case: $f=(b+x)^3-b^3=x3b^2+x^2 3b+x^3$
In the 4-dimensional case: $f=(b+x)^4-b^4=x4b^3+x^2 4b^2+x^3 4b+x^4 + e$ with $e=2b^2x^2$
In the 5-dimensional case: $f=(b+x)^5-b^5=x5b^4+x^2 5b^3+x^3 5b^2+x^4 5b+x^5 + e$ with $e=5b^3x^2+5b^2x^3$
In the 6-dimensional case: $f=6 b^5 x + 6 b^4 x^2 + 6 b^3 x^3 + 6 b^2 x^4 + 6 b x^5 + x^6 + e$ with $e=9 b^4 x^2 + 14 b^3 x^3 + 9 b^2 x^4$
This should be generally represented as: $$f=x^n + n \sum_{i=1}^{n-1}{b^i x^{(n-1)+1-i}} + e(x,n)$$
Now I am searching for the correct equation of $e(x,n)$.
We can compactly express the whole (general) formula as
$$f = \sum_{i=1}^{n}\binom{n}{i} \cdot b^{n-i} \cdot x^i$$
so that $$f = \sum_{i=1}^{n}\frac{n!}{i!\cdot(n-i)!} \cdot b^{n-i} \cdot x^i$$ is a very nice way to express your relation, at the end.