Does $a^nb^n=(ab)^n$ apply to $c^m(x-x_0)^m$?

Question

Does the power of a product rule

$$a^nb^n=(ab)^n, a,b \in \mathbb{R}, n \in \mathbb{N}$$ apply to $$c^m(x-x_0)^m, c,x, x_0 \in \mathbb{R}, m \in \mathbb{N}$$

so that

$$c^m(x-x_0)^m=(cx-cx_0)^m$$

?

Answer 1

Sure it does! Since $a,b\in\mathbb R$ and $n\in\mathbb N$ is really all required to let $a^nb^n=(ab)^n$ hold, we merely need to have $c^m(x-x_0)^m$ to have the same form and fit all the requirements.

Simply allow the substitution $a=c$, $b=x-x_0$, and $n=m$ to have

$$a^nb^n=(ab)^n\implies c^m(x-x_0)^m=(c(x-x_0))^m$$