If $f(z)=\sum a_nz^n$, what is $\sum n^3a_nz^n$?
The desired sum is $a_1z+8a_2z^2+27a_3z^3+\cdots$. I can't see how to write the desired sum in terms of $f$. For example, I could substitute $kz$ for $z$ to get $f(kz)=\sum k^na_nz^n$, but that doesn't help me get the coefficient $n^3$.
Hint: Differentiate, multiply by $z$, differentiate, $\dots$.
If $f(z)=\sum a_nz^n$, then $zf'(z) = \sum n a_n z^n$. Applying this thrice you would get-
$\sum n^3 a_n z^n = z \left(f'(z) + 3z f''(z) + z^2 f'''(z)\right)$
Find the coordinates $(\alpha_0,\alpha_1,\alpha_2,\alpha_3$) of $x^3$ in the basis $(1,x,x(x-1),x(x-1)(x-2))$ of $\mathbb R_3[x]$ so $$\sum n^3a_n z^n=\alpha_0f(z)+\alpha_1 z f'(z)+\alpha_2 z^2 f''(z)+\alpha_3z^3f'''(z)$$
Hint: $$z(a_0 + a_1 z + a_2 z^2 + \ldots)' = a_1 z + 2 a_2 z^2 + \ldots$$
For the general case, you will encounter the Stirling numbers of the second kind.
In this case, you are given $f(z)=\sum_{n=0}^{\infty} a_nz^n$ and want to find $f_k(z) =\sum_{n=0}^{\infty} n^ka_nz^n $.
By repeated differentiation, $f^{(k)}(z)=\sum_{n=k}^{\infty} (n)_k a_nz^{n-k}$ where $(n)_k =n(n-1)...(n-k+1) $ is the Pochhammer symbol denoting the falling factorial.
Therefore $\sum_{n=0}^{\infty} (n)_k a_nz^{n} =z^kf^{(k)}(z) $.
We now use the definition of the Stirling numbers of the second kind: $\sum_{j=0}^k S(k, j) (n)_j =n^k $.
Note: The Stirling numbers of the first kind go the other way, expressing $(n)_k =\sum_{j=0}^k s(k, j) n^j $
These make is easy to get a formula for $f_k(z)$:
$\begin{align} f_k(z) &=\sum_{n=0}^{\infty} n^ka_nz^n\\ &=\sum_{n=0}^{\infty} a_nz^n\sum_{j=0}^k S(k, j) (n)_j\\ &=\sum_{j=0}^k\sum_{n=0}^{\infty} a_nz^n S(k, j) (n)_j\\ &=\sum_{j=0}^k S(k, j) \sum_{n=0}^{\infty} a_nz^n (n)_j\\ &=\sum_{j=0}^k S(k, j) z^k f^{(k)}(z)\\ \end{align} $
For $k=3$, $S(3, 1..3) = [1, 3, 1]$, so $\sum_{n=0}^{\infty} n^3a_nz^n =z^3 f'''(z)+3z^2f''(z)+zf'(z) $.
For $k=4$, $S(4, 1..4) = [1, 7, 6, 1]$, so $\sum_{n=0}^{\infty} n^4a_nz^n =z^4 f^{(4)}(z)+7z^3 f'''(z)+6z^2f''(z)+zf'(z) $.
