Consider the general case, $n^{\text{th}}$ root of real $x$ to the integer power $p$: $\sqrt[n]{x^p}$ When determining the simplest form, find the quotient $q$ and remainder $r$ when $p$ is divided by $n$. $p=n\cdot q+r$ The simplest form is $\sqrt[n]{x^p}=\begin{cases}\left\vert{x^q}\right\vert\sqrt[n]{x^r}&\text{$n$ even, $p$ even, $q$ odd, and $r$ even}\\\text{or}\\x^q\sqrt[n]{x^r}&\text{any other case, or $x$ non-negative}\end{cases}$ The reason the absolute value is only required in such a specific case is because if - if $n$ is odd, then the result is allowed to be negative - if $p$ is odd (but $n$ is even), then the domain only contains non-negative $x$ - if $q$ is even, then $x^q$ is already non-negative - if $r$ is odd, then the domain only contains non-negative $x$