Cubed Root of an Expectation value $\left<\frac 1 {r^3}\right>$ I am afraid this question will reveal how little I know about expectation values but it cannot be helped. Say I have the expectation value:
$$\left<\frac 1 {r^3}\right> = \frac {1}{(na_0)^3} \frac {2}{\ell(\ell+1)(2\ell+1)}$$
If I know the values of $n, a_0$ and $\ell$ and I wanted to find $\left<\frac {1}{r}\right>$ could I simply take the cubed root of the value of $\left<\frac 1 {r^3}\right>$?
 A: Generally speaking no. A counter example is the standard deviation:
$$\sigma_X^2=\langle X^2\rangle-\langle X\rangle^2$$
The fact that this is an important quantity indicates that generally you can't take powers outside an expectation value, otherwise the standard deviation would be zero. In some examples you might be able to take powers outside of an expectation value but without further knowledge of your system it's impossible to tell.
Edit: to make this more explicit compare
\begin{align}
\langle X^n\rangle&=p_1 X_1^n+\dots+p_NX_N^n\\
\langle X\rangle^n&=\left(p_1 X_1+\dots+p_NX_N\right)^n
\end{align}
A: You cannot do this in general.  Consider for instance the operator $\hat x$ for a simple harmonic oscillator.  For an eigenstate of $\hat H$, we have $\langle x\rangle=0$ since $x$ can take positive and negative values, and is equally probable on the left and the right of $x=0$.  On the other hand, $\hat x^2$ can only take positive values so that $\langle x^2\rangle\ge 0$ necessarily.
A: No you can't.
According to Expectation powers of $r$ for hydrogen
you have the expection values
$$\left<\frac{1}{r^3}\right> = \frac {1}{(na_0)^3} \frac {2}{\ell(\ell+1)(2\ell+1)}$$
and
$$\left<\frac{1}{r}\right> = \frac {1}{n^2a_0}$$
You can clearly see:
$$\left<\frac{1}{r^3}\right> \ne \left<\frac{1}{r}\right>^3$$
Actually this inequality is more general.
For any operator $\hat{A}$ you usually have:
$$\left<A^n\right> \ne \left<A\right>^n$$
However, there is an important special case where they are equal:
If the state vector $|\psi\rangle$ happens to be an eigenvector of $\hat{A}$, i.e.
$$\hat{A}|\psi\rangle = a|\psi\rangle$$
where $a$ is just a number (the so-called eigenvalue).
Then it is
$$\left<A^n\right> = \left<A\right>^n.$$
You can easily prove this by using the definition of the expection
value $\langle\hat{A}\rangle=\langle\psi|\hat{A}|\psi\rangle$,
and the normalization condition $\langle\psi|\psi\rangle=1$.
Applied to your case $\hat{A}=\frac{1}{r}$ this means your wave function
would need to look like this:

Here $\psi(r)$ has an infinitesimally narrow peak at some special radius $r_0$ and is zero everywhere else. Only then you have
$$\left<\frac{1}{r^3}\right> = \left<\frac{1}{r}\right>^3 = \frac{1}{r_0^3}$$
