Question: How would you prove the following formulas:
\begin{align*} & a^n-b^n=(a-b)\sum_{k=1}^na^{n-k}b^{k-1}\qquad\qquad\qquad[\text{Always}]\\ & a^n-b^n=(a+b)\sum_{k=1}^n-(-1)^ka^{n-k}b^{k-1}\qquad[\text{If n is even}]\\ & a^n+b^n=(a+b)\sum_{k=1}^n-(-1)^ka^{n-k}b^{k-1}\qquad[\text{If n is odd}]\end{align*}
I find these formulas incredibly interesting and wonder how they came up with them in the first place. I don't know where to begin in the proof.
Note that I do understand that you can multiply out the right hand side and see if both sides give an identity, but I don' really consider that a proof because it doesn't tell me how the person got to this formula in the first place.
I do realize that the third formula is the same as the second formula for when $b=-b$. So if I can prove the first and second formulas, then simply replacing $b$ with its negated version wil give you formula $3$.
These probably were discovered, like many results, by looking at results for small $n$ and seeing how they could be generalized.
For example, looking at $a^2-b^2 = (a-b)(a+b)$ and $a^3-b^3 = (a-b)(a^2+ab+b^2) $ would suggest the first formula.
$a^3+b^3 =(a+b)(a^2-ab+b^2) $ would suggest the third formula.
As for your second formula, it is wrong. If we put $n=2$ it says $a^2-b^2 =(a-b)(-\sum_{k=1}^2 (-1)^ka^{2-k}b^{k-1}) =(a-b)(-(-a+b)) =(a-b)(a-b) $ which is false.