# $x^q - 1 = (x - 1) \cdot (?)$ for $q \in \mathbb R_{>0}$

As we know, for $$q \in \mathbb N$$ we have $$x^q - 1 = (x-1) \cdot (x^{q-1} + x^{q-2} + \cdots + 1)$$.

Does some kind of similar expansion exist for $$q \in \mathbb R_{>0}$$? I'm not sure, but at the very least

$$\lim_{x \to 1} \frac {x^q - 1} {x - 1} = \lim_{x \to 1} q x^{q-1} = q,$$

so it would make some sense to separate $$x-1$$ from everything else.

My only thought is to use Taylor expansion for $$(1 + t)^q$$ (where $$t = x - 1$$):

$$\frac {(1 + t)^q - 1} {t},$$

and it'll find expression as a power series of $$t$$. But this power series

• Doesn't always converge (while the initial expression is defined everywhere)
• Is in terms of $$t=x-1$$. It's unclear to me how to convert it into power series of $$x$$ so that coefficients have a reasonable representation.