Show that $\gcd\left(\frac{a^n-b^n}{a-b},a-b\right)=\gcd(n d^{n-1},a-b)$

Question

How to show that

$$ \gcd\bigg( {a^n-b^n \over a-b} ,a-b\bigg )=\gcd(n d^{n-1},a-b ) $$ $a,b\in \mathbb Z$

where $d=\gcd(a,b)$?

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Note $\ $ Some of the answers below were merged from this question. The answers (and their comments) may depend on context provided in that question.

Answer 1

We have $\large\ d=(a,b)\ ,\ $ thus $\large\ \exists\ A,B\ \ \ a=Ad,\ b=Bd,\ (A,B)=1$

$\large\left(\LARGE\frac{a^n-b^n}{a-b}\large,a-b\right)=(n d^{n-1},a-b)$
$\large\ d\left(d^{n-2}\cdot\LARGE\frac{A^{\ n}-B^{\ n}}{A-B}\large,A-B\right)=d(n d^{n-2},A-B)$

Let $\large\ m=A-B\ ,\ \ \ \ $ then $\large\ (m,B)=1$

$\large\ \left(d^{n-2}\cdot\LARGE\frac{(B+m)^n-B^n}{m}\large,m\right)=(nd^{n-2},m)$
$\large\ \left(d^{n-2}\cdot(nB^{n-1}+Qm),m\right)=(nd^{n-2},m)\ \ \ \ $ for some integer Q
$\large\ \left(nd^{n-2}B^{n-1},m\right)=(nd^{n-2},m)\ ,\ \ $ which is due to $\large (m,B)=1$.

Answer 2

Putting $c=a-b,$ we get, $$(a-b, \frac{a^n-b^n}{a-b})=(c, \frac{(b+c)^n-b^n}c)=(c,\binom n 1 b^{n-1}+\binom n 2 b^{n-2}c+\cdots+c^{n-1})=(c,nb^{n-1})$$

As $(c,b)=(a-b,b)=(a,b)=d,$ let $\frac c C=\frac b B=d$ so that $(B,C)=1$

$$(c,nb^{n-1})=(Cd,nB^{n-1}d^{n-1})=d(C,nB^{n-1}d^{n-2})=d(C,nd^{n-2})$$ as $(B,C)=1$

$$(c,nb^{n-1})=d(C,nd^{n-2})=(Cd,nd^{n-1})=(c,nd^{n-1})=(a-b, nd^{n-1})$$