What is the generating function for $(a_n)_{0\leq n}$, where $a_n$ is defined as $a_0 = 0$ and $a_{n+1} = a_n + (n+1)^3$ for all $n\geq0$? I have an idea of how to approach this problem, but I am rather confused by generating functions so I do not know if some of the steps I take are correct/allowed. I started this by noting that the function we are looking for is $f(x) = \sum_{n\geq0}a_nx^n$.
Since the sequence can be defined recursively by $a_0 = 0$ and $a_{n+1} = a_n + (n+1)^3$ for all $n\geq0$, I multiplied each of the parts by $x^n$ and summed to get
$$\sum_{n\geq0}a_{n+1}x^n = \sum_{n\geq0}a_nx^n + \sum_{n\geq0}(n+1)^3x^n$$
Which I rewrote as
$$\frac{1}{x}\sum_{n\geq0}a_{n+1}x^{n+1} = f(x) + \sum_{n\geq0}(n+1)^3x^n$$
And, since $a_0 = 0$, 
$$\frac{1}{x}f(x) = f(x) + \sum_{n\geq0}(n+1)^3x^n$$
I then subtracted $f(x)$ from both sides to obtain
$$\frac{1-x}{x}f(x) = \sum_{n\geq0}(n+1)^3x^n$$
This is where I got stuck. I think the next step would be to say, since $0^3 = 0$, we can rewrite it as
$$\frac{1-x}{x}f(x) = \sum_{n\geq0}n^3x^n$$
I am not sure if this works/is true? In a previous problem, we derived the generating function for $(n^3)$, so I subsituted it in here
$$\frac{1-x}{x}f(x) = \frac{x^3+4x^2+x}{(1-x)^4}$$
And lastly multiplied both sides by $\frac{x}{1-x}$ to obtain
$$f(x) = \frac{x^4+4x^3+x^2}{(1-x)^5}$$
So, does this look right? Is the maths correct? I am still new to generating functions so I am still confused by a lot of the rules and what you can and cannot do.
 A: (A lot of people wrote long answers with alternate solutions and ideas -- I am suspicious some may  not have actually read your post. Probably it will be more helpful to you to get direct feedback on what you did, so I will do that.)
It's almost a fully correct solution.
The only error is when you replaced $\sum_{n \ge 0} (n+1)^3 x^n$ with $\sum_{n \ge 0} n^3 x^n$.

I am not sure if this works/is true?

It doesn't quite. Let's do it more carefully, by changing the index $n$ to $n' = n+1$ (so conversely, $n = n' - 1$):
\begin{align*}
\sum_{n \ge 0} (n+1)^3 x^n
  &= \sum_{\color{red}{n' \ge 1}} (\color{red}{n'})^3 x^{\color{red}{n' - 1}} \\
  &= \frac{1}{x} \sum_{n' \ge 1} (n')^3 x^{n'} \\
  &= \frac{1}{x} \sum_{n' \ge 0} (n')^3 x^{n'} \quad (\text{since } 0^3 = 0)\\
  &= \frac{1}{x} \sum_{n \ge 0} n^3 x^n \quad \text{(rename the variable)}
\end{align*}
OK, so you'll end up with the same thing as before, but with an additional $\frac{1}{x}$ factor.
A: I'll post an incomplete solution and leave it to you to finish it:
$$\begin{align*}
\color{blue}{\sum_{n>0}a_nx^n}&=x\sum_{n\ge0}a_{n+1}x^n\\
&=x\sum_{n\ge0}(a_n+(n+1)^3)x^n\\
&=x\left(\color{blue}{\sum_{n>0}a_n x^n}+\sum_{n\ge0}(n+1)^3x^n\right)\\
\end{align*}$$
Take particular note of the indexing changes I did, which exploit the initial condition $a_0=0$.
You should be able to evaluate one of the sums explicitly, and then solve for the unknown sum.
A: The definition of generating functions is very clear $f(x)=\sum\limits_{n=0}^{\infty}a_nx^n$. I prefer not ignoring $a_0$ to avoid messing with the indexes and, typically, you calculate it from the initial conditions, which you mentioned only in the title. So, I will treat it as unknown parameter for the purpose of the calculations below.
Thus
$$f(x)=\sum\limits_{n=0}^{\infty}a_nx^n=
a_0+\sum\limits_{n=1}^{\infty}a_nx^n=
a_0+\sum\limits_{n=1}^{\infty}(a_{n-1}+n^3)x^n=\\
a_0+\sum\limits_{n=1}^{\infty}a_{n-1}x^n+\sum\limits_{n=1}^{\infty}n^3x^n=
a_0+x\sum\limits_{n=1}^{\infty}a_{n-1}x^{n-1}+\sum\limits_{n=1}^{\infty}n^3x^n=\\
a_0+x\sum\limits_{n=0}^{\infty}a_{n}x^{n}+\sum\limits_{n=1}^{\infty}n^3x^n=a_0+xf(x)+x\sum\limits_{n=1}^{\infty}n^3x^{n-1}=\\
a_0+xf(x)+x\left(\sum\limits_{n=1}^{\infty}n^2x^{n}\right)'=
a_0+xf(x)+x\left(x\sum\limits_{n=1}^{\infty}n^2x^{n-1}\right)'=\\
a_0+xf(x)+x\left(x\left(\sum\limits_{n=1}^{\infty}nx^{n}\right)'\right)'=
a_0+xf(x)+x\left(x\left(x\sum\limits_{n=0}^{\infty}(n+1)x^{n}\right)'\right)'=\\
a_0+xf(x)+x\left(x\left(\frac{x}{(1-x)^2} \right)'\right)'=
a_0+xf(x)+\frac{x^3+4x^2+x}{(1-x)^4}$$
and 
$$f(x)=\frac{a_0}{1-x}+\frac{x^3+4x^2+x}{(1-x)^5}$$
