What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$ Here's what I tried:
$$\sum_{n \ge0} {\frac{1}{3} \choose n} x^n= \sum_{n \ge0} = \frac{\frac{1}{3}!}{n!(n-\frac{1}{3})!}x^n=\sum_{n \ge0} \frac{(\frac{1}{3}-1)(\frac{1}{3}-2)\cdot ...\cdot(\frac{1}{3}-(n-1)) }{n!}x^n$$
What to do more, or is this all wrong?

Hint: Supplement to the already given answer
If a series
\begin{align*}
A(x)=\sum_{n\geq 0}a_nx^n
\end{align*}
is stated, the coefficient of $x^k$ is $a_k$ without any notion of $x$. A convenient notation for the coefficient of $x^k$ is $[x^k]$.

Since the coefficient of $x^4$ should be obtained, an answer could be stated as
\begin{align*}
[x^4]\sqrt[3]{1+x}&=[x^4]\sum_{n \ge0} {\frac{1}{3} \choose n}x^n\\
&=\binom{\frac{1}{3}}{4}\\
&=\frac{1}{4!}\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)\\
&=-\frac{10}{243}
\end{align*}

