Significance of $\sqrt[n]{a^n} $?!

There is a formula given in my module:

$$ \sqrt[n]{a^n} = \begin{cases} \, a &\text{ if $n$ is odd } \\ |a| &\text{ if $n$ is even } \end{cases} $$

I don't really understand the differences between them, kindly explain with an example.

$\sqrt{(-1)^2} =\sqrt{1} = 1 = |-1|$ because in order to make the square root an unambiguous operation, we agree that the square root of a nonnegative number $x$ is always the (unique) nonnegative number $r$ such that $r^2=x$. But with cubic roots there is no problem: $\sqrt[3]{-1}=-1$, because every real number has a unique cubic root.

The same is true with 4th, 6th, 8th, 10th, etc. powers, since $a^n = (-a)^n$, and the 4th, 6th, 8th, 10th, etc roots are defined to be the unique nonnegative real number that "works", so that they are unambiguous.

That is, there are two numbers which when squared will given you the value $2^2$: both $2$ and $-2$. There are two number that when taken to the fourth power will give you $(-6)^4$: both $-6$ and $6$. And so on. Generally, both $a$ and $-a$ will, when raised to an even $n$th power, give the same answer: $a^n = (-a)^n$. And we agree that a square root (fourth root, sixth root, etc) will always be the nonnegative answer, so the $n$th root of $a^n$ will be $|a|$ when $n$ is even. (Don't let the big $-$ in "$-a$" fool you; that does not mean that $-a$ is negative, it just means the additive inverse of whatever $a$ is; if $a$ is positive, then $-a$ is negative, but if $a$ is negative, say $a=-3$, then $-a$ is positive, $-a = -(-3) = 3$. Repeat after me: the proper way to pronounce "-a" is not "negative a", the proper pronunciation is "minus a").

But if $n$ is odd, then every number has a unique $n$th root. In particular, the only number that when cubed gives $2^3$ is $2$; the only number which, when raised to the fifth power, gives $(-6)^5$, is $-6$. There is no longer the problem that both $6$ and $-6$ are possible answers, so we can simply say that the cubic root of $(-2)^3$ is $-2$, the fifth root of $7^5$ is $7$, etc.

The comments for this answer have more value than the answer I posted. I am removing the contents of my answer but leaving the responses in tact ( that is why I am not deleting this post, if there is an alternative method please let me know). Regarads