Can you explain why $(\sqrt{n})^3 = \sqrt{n^3}$? That this, why does, for example, the square root of $n$, cubed, give the same value as the square root of $n$ cubed?
 A: The reason that taking a square root fundamentally "makes sense" (i.e. is well-defined) as a function is as follows:

For any $x \geq 0$: if $a,b \geq 0$ are such that $a^2 = x$ and $b^2 = x$, then $a = b$

Now, let's look at those two values you have: that is, take $a = (\sqrt{n})^3$ and $b = \sqrt{n^3}$.  We note that
$$
a^2 = (\sqrt{n})^3 \cdot (\sqrt{n})^3 = (\sqrt{n})^6 = ((\sqrt{n})^2)^3 = n^3\\
b^2 = \sqrt{n^3} \cdot \sqrt{n^3} = \sqrt{n^3 \cdot n^3} = \sqrt{(n^{3})^2} = n^3
$$
since $a,b$ are both non-negative and $a^2 = b^2$, it must be that $a = b$.

Here's an equivalent perspective: if $p$ and $q$ are integers, then $(n^{p})^q = (n^q)^p$.  This is because, with the way exponentiation is defined, we have
$$
(n^q)^p = \left(\overbrace{n \cdots n}^{p \text{ times}}\right)^q = 
\overbrace{\overbrace{n \cdots n}^{p \text{ times}} \cdots \overbrace{n \cdots n}^{p \text{ times}}}^{q \text{ times}} = 
\overbrace{n \cdots n}^{pq \text{ times}} = n^{pq}\\
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(n^q)^p = \left(\overbrace{n \cdots n}^{q \text{ times}}\right)^p = 
\overbrace{\overbrace{n \cdots n}^{q \text{ times}} \cdots \overbrace{n \cdots n}^{q \text{ times}}}^{p \text{ times}} = 
\overbrace{n \cdots n}^{qp \text{ times}} = n^{qp}
$$
and of course, $pq = qp$.  When we extend the definition of exponentiation to rational numbers $p$ and $q$, the new exponents inherit the property $(n^p)^q = (n^q)^p = n^{pq}$.  Thus, we have
$$
(\sqrt{n})^3 = (n^{1/2})^3 = n^{3/2} = (n^3)^{1/2} = \sqrt{n^3}
$$
as desired.
A: can you see why $(x^3)^2=x^6=(x^2)^3$?
If you do you can use this to show both of these non-negative numbers give the same result when raised to the sixth power. 
A: $\sqrt{n^3}$ is the positive, real number $x$ such that $x^2=n^3$. And $\sqrt n$ is the positive, real number $y$ such that $y^2=n$. But
$$(y^3)^2=(y^2)^3=n^3$$
This shows that $y^3=x$.
