Factor $d^k+(a-d)^k$

Question

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?

Answer 1

Start by understanding how to factor $$ x^n - y^n . $$

Presumably you know how to do that when $n=2$. For $n=3$ you can check that $$ x^3 - y^3 = (x-y)(x^2 + xy + y^2) $$

Now guess for higher powers.

Then see what happens if $n$ is odd and you replace $y$ by $(-y)$.

Answer 2

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.

Answer 3

They used that

$$x^k+y^k=(x+y)(x^{k-1}-x^{k-2}y- \dots +xy^{k-2}+y^{k-1})$$

holds for $k$ - odd and have plugged $x=d$, $y=a-d$