From my old high school math textbook:
If ${a{\geq }0}$ and $n\in \mathbb{N} ^{\ast }$, then ${\sqrt[{n}] {a}}$ is the non-negative solution of ${{x}^{n}}=a$.
It then goes on to infer a number of properties and identities in the case that $\alpha \geq 0$ and $n\in \mathbb{N} ^{\ast }$.
However the choice of defining the n-th root of something only as the positive solution of this equation leaves me quite confused. For example, we could define $\sqrt[3] {-8}=-2$ but we don't.
Instead the book goes on to break down the solution of ${{x}^{n}}=a$ based on $\pm \sqrt[n] {\left| a\right| }$ instead of just $\sqrt[n] {a}$.
Why do we do this?
The second part is please tell me if these expressions are in fact correct because the book doesn't mention them (although it might be using them) and I've seen them elsewhere online.
If $x \geq 0$ and $a\geq 0$ and $n\in \mathbb{N} ^{\ast }$, then $\ x^{n}=a\Leftrightarrow x=\sqrt[n] {a}$
$\left( \sqrt[n] {a}\right) ^{n}=a$ holds for every $a\in \mathbb{R}$, while $\sqrt[n] {a^{n}}$ holds only for $a\geq 0$
It is quite likely that your textbook deals with it that way in order not to have to deal with case case in which $n$ is even (in which case no negative number has a $n$th root) and the case in which $n$ is odd (in which case each number has one and only one $n$th root) separately.
And, yes, those assertions are correct.