How to expand $a^n + b^n$ for all positive exponents?

This question is similar to: How to simplify $a^n - b^n$?
where the accepted answer is given as: $a^n-b^n=(a-b)\Big(\sum_{i=0}^{n-1}a^{n-1-i}b^i\Big)$

Also, when n is odd, then: $a^n + b^n = (a+b)\Big(\sum_{i=0}^{n-1}(-1)^ia^{n-1-i}b^i\Big)$ but the solution is limited to odd exponents whereas the solution above for $a^n -b^n$ accounts for all positive exponents.

Is there a solution for $a^n + b^n$ similar to $a^n -b^n$ that accounts for all positive exponents ?

I suggest \begin{equation} a^n + b^n = \prod_{k=0}^{n-1}\left(a - b\, e^{\frac{i\pi}{n}(2k+1)}\right) \end{equation}

If $n=st$ with $s>1$ odd, then we have $$ a^n+b^n=(a^t)^s+(b^t)^s $$ and you can use your "$n$ is odd" factorisation with $s$ instead.

If $n$ is a power of $2$, there is no general method that I know of. Which is evidenced, for instance, by the fact that the Fermat numbers $2^{2^k}+1^{2^k}$ are sometimes prime, and it is unknown how many times that happens. We know it happens at least 5 times (for $k=0,1,2,3$ and $4$), and there are believed to be no more.