Is it possible to prove the formula to factorize $(a^n + b^n)$ using the formula that holds for $(a^n - b^n) $? ( with $n\geq 3$)

Question

Is it possible to prove the factorization of $a^3 + b^3$ such that $a^3 + b^3= (a+b) ( a^2 - ab + b^2)$ using the formula for the factorization of

$ (a^n - b^n) = (a-b) ( a^{n-1} + a^{n-2}b^1 + ... + ... + a^1 b^{n-2} + b^{n-1}) $ ,

and , in particular the instance of this formula with $n=3$, I mean :

$ a^3 - b^3 = (a-b) ( a^2 + ab + b^2) ?$

The problem I see is this : certainly, $a^n + b^n = a^n - (-b^n)$. But , in order to apply the formula that holds for the difference , shouldn't we have also : $ a^n - (-b)^n$?

If I'm right in asserting this necessary condition, the question arises : is it the case in general that

$a^n - (-b^n)=a^n - (-b)^n$ ?

More generally, how to prove the " sum of $n^{\mathbb {th}}$ powers" formula?

Answer 1

Just let $c=-b$. Then $a^3+b^3=a^3-c^3=(a-c)(a^2+ac+c^2)=(a+b)(a^2-ab+b^2)$.

Note that this only works because $3$ is odd.