Factor $d^k+(a-d)^k$ I was reading a number theory book and it was stated that $d^k+(a-d)^k=a[d^{k-1}-d^{k-2}(a-d)+ . . .+(a-d)^{k-1}]$ for $k$ odd. How did they arrive at this factorization? Is there an easy way to see it?

It's slightly easier if we switch to $x=d,\,y=a-d$ so you want to verify $x^k+y^k=(x+y)(x^{k-1}-x^{k-2}+\cdots +y^{k-1})$ for odd $k$. You can now easily prove this by $k\mapsto k+2$ induction, or by taking it as the odd-$k$, $y=-z$ special case of $x^k-z^k=(x-z)\sum_{j=0}^{k-1}x^j z^{k-1-j}$. If you just want to prove divisibility without worrying about the quotient, you can also work modulo $x+y$, viz. $x=-y\implies x^k=(-1)^k y^k$. The power of $-1$ simplifies to $-1$ as desired for $k$ odd.
