Interpretation of $a+b \ | \ a^n + b^n$ for odd $n$

Question

It is not hard to show that $a+b \ | \ a^n + b^n$ for odd $n$.

(because $f(x) = x^n - b^n = (x-b)h(x)$ we have $a - b \ | \ a^n - b^n$, so $a - (-b) \ | \ a^n - (-1)^n b^n$)

Is there a nice interpretation of this fact? (Or if not for all odd $n$, at least for $n = 3$ or $n = 5$?)

Answer 1

As you mentioned in the last comment you may see $a^3+b^3=(a+b)(a^2-ab+b^2)$

You may have a general view of this as $a^n+b^n$ has always $(a+b)$ as a factor.

You can prove this by binomial expansion.

Answer 2

I have some interpretation of this fact, for $n = 3$ (and also other $n$ if you are willing to imagine $n$-cubes)

Because $f(x) = x^3 + a^3 = (x + a)(x + \zeta_3 a)(x + \zeta_3^2 a)$ we get that $a^3 + b^3 = (a+b)(\zeta_3 a+ b)(\zeta_3^2 a + b)$ where $\zeta_3 = -\frac{1}{2}-\frac{i \sqrt{3}}{2}$.

We can plot these in $\mathbb{C}$ in the following way:

If we now rotate the vector $\zeta_3 a + b$ such that it aligns with the $i\mathbb{R}$ axis, and we rotate $\zeta_3^2 a + b$ such that it aligns with the axis perpendicular to the plane and then complete the cube, we get a cube with volume $a^3 + b^3$ with one rib being $a+b$.

It is possible to compute these rotated vectors, although too time consuming for me at this moment.