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Would anyone kindly present a clear and concise proof for the following identities?

Given two real numbers x, y and a positive integer n,

$$x^n-y^n=(x-y)\times\sum_{k=0}^{n-1}x^{(n-1)-k}y^k$$

Similarly, if n is odd,

$$x^n+y^n=(x+y)\times\sum_{k=0}^{n-1}(-1)^kx^{(n-1)-k}y^k$$

Any interesting generalisations and proofs for said generalisations are welcome as well.

Albert
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  • If you have to discover such an identity (in the sense you are asked to find a polynomial $p$ in $x$ and $y$ such that $x^{n}-y^{n}=(x-y)p(x,y)$ you can think of $y$ as fixed and use the standard process of long division. Do you know how to diivide one polynmial by another? – Kavi Rama Murthy Jul 27 '19 at 11:41
  • Why don't you simply figure out what happens on multiplying the two sums – Anand Jul 27 '19 at 11:45
  • start with $(x-y)(x^2+xy+y^2)=$ $x^3+x^2y+xy^2-yx^2-yxy-y^3=$ $x^3-y^3$. Note, if $n$ is odd, then $x^3+y^3=$ $x^3-(-y^3)=$ $x^3-(-y)^3=$ $(x-(-y))(x^2+x(-y)+(-y)^2)=$ $(x+y)(x^2-xy+y^2)=$ – Mirko Jul 27 '19 at 15:10

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Start on the right and be straight-forward: $$\begin{align} (x-y)\times\sum_{k=0}^{n-1}x^{(n-1)-k}y^{k}&=x\sum_{k=0}^{n-1}x^{(n-1)-k}y^{k}-y\sum_{k=0}^{n-1}x^{(n-1)-k}y^{k}\\ &=\sum_{k=0}^{n-1}x^{n-k}y^{k}-\sum_{k=0}^{n-1}x^{(n-1)-k}y^{k+1} \\ &=\sum_{k=0}^{n-1}x^{n-k}y^{k}-\sum_{k=1}^{n}x^{n-k}y^{k}\\ &=x^n+\sum_{k=1}^{n-1}x^{n-k}y^{k} -\sum_{k=1}^{n-1}x^{n-k}y^{k}-y^n\\ &=x^n-y^n.\end{align}$$

For the second, substitute $-y$ for $y$.

Hagen von Eitzen
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