What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$

Question

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?

Answer 1

The fourth term should be:

$\dfrac{\left(\dfrac{1}{3}\right)\left(-\dfrac{2}{3}\right)\left(-\dfrac{5}{3}\right)\left(-\dfrac{8}{3}\right)}{4!}x^4 $

Which should come out to:

$-\dfrac{10}{243}x^4 $

Answer 2

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*}