Show that if $\gcd(a,b)=1$ and $p$ is an odd prime, then ${\gcd(a+b,}\frac{a^p +b^p}{a+b}$$) = 1$ or $p$
Sorry about the duplicate
In another answer, however, the sum $\sum\limits_{k=0}^{n-1} (-1)^{k}a^{n-1-k}b^{k}$ was expressed as $\left(\sum\limits_{k=0}^{n-2} (-1)^{k}(k+1)a^{n-2-k}b^{k}\right)(a+b) + (-1)^{n-1}nb^{n-1}$
How was it done?
On
Hint $\ $ If $\,f\in\Bbb Z[x]\,$ then the Factor Theorem yields a Taylor approximant
$\quad\qquad\qquad\quad\, f(x) = f(a) +\: f\:'(a)\ (x-a) \:+\: (x-a)^2\: g(x)\quad {\rm for\ some}\ \ g(x) \in \mathbb Z[x]$
$\displaystyle\qquad\qquad\Rightarrow\ \ \frac{f(x)-f(a)}{x-a} \: \equiv\ f\:'(a)\quad ({\rm mod}\ \:x-a)$
$\quad \Rightarrow \left(x-a,\dfrac{f(x)-f(x)}{x-a}\right)\,=\, (x-a,\, f'(a))\ $ by the Euclidean algorithm
$\ \ \underset{\large \begin{array}{}\ \ x\,=\,b\\ a\ \to\ -a\end{array}}{\overset{\large f(x)\,=\,x^p}\Rightarrow} \left(b+a,\dfrac{b^p+a^p}{b+a\ }\right)\,=\!\! \underbrace{(b+a,\, pb^{p-1})}_{\large (b+a,b)\,=\,(a,b)\,=\,1}\!\! = (b+a,p) = 1\,\ {\rm or}\,\ p$
Hint:
$$a^p = ((a+b)-b)^p = \sum_{k=0}^p (-1)^k\binom{p}{k} (a+b)^{p-k}b^k.$$