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?

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 $
