General factorization of $x^n-y^n$ (n is real)

Question

I am looking for a general expansion of $x^{n}-y^{n}$ with $x,y>0$ and $n$ being real. I came across the following formula (Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$)

$$x^n-y^n = (x-y)(x^{(n-1)}+x^{(n-2)}y+...+y^n)$$

Does it hold true for $n \in \mathbb{R}$?

Answer 1

If you write $x = y + t$, then you can use the binomial series on $x^n = (y+t)^n$, so

$$\eqalign{ x^n - y^n &= \sum_{k=1}^\infty {n \choose k} y^{n-k} t^k\cr &=\sum_{k=1}^\infty {n \choose k} y^{n-k} (x-y)^k\cr &= (x-y) \sum_{k=1}^\infty {n \choose k} y^{n-k} (x-y)^{k-1}} $$ The series converges for $|x-y| < |y|$.

Answer 2

You could factor out $x - y$, giving you:

$$(x - y)(x^{n-1} + x^{n-2}y + ... xy^{n - 2} + y^{n - 1}) = \\(x - y)\sum_{i=0}^{n-1}x^{i}~y^{n - 1 - i}$$

Edit: The question was changed to include this factorization after this answer was posted.