What is the value of $\sqrt[n]{(x-1)^n}$?

Question

Let $n \in \mathbb{N}$. If $n=2$, it is known that $\sqrt{(x-1)^2}=|x-1|$. If, however, the value of $n$ is unknown what is the result of $\sqrt[n]{(x-1)^n}$?

1. $\sqrt[n]{(x-1)^n}=x-1$ or
2. $\sqrt[n]{(x-1)^n}=|x-1|$?

Answer 1

Since $n$ is unknown, all you can say is that$$\sqrt[n]{(x-1)^n}=\begin{cases}|x-1|&\text{ if $n$ is even}\\x-1&\text{ if $n$ is odd.}\end{cases}$$

Answer 2

If you want a formula that is not casewise on the parity of $n$ (or rather, has the appearance of not being casewise), then $$\sqrt[n]{(x-1)^n}=(x-1)\cdot(-1)^{(n+1)\cdot\left((1-\operatorname{sign}(x-1))!-1\right)}$$

or try

$$\sqrt[n]{(x-1)^n}=(x-1)\cdot(-1)^{(n+1)\cdot2^{\operatorname{sign}(x-1)+1}}$$