Answer
$x-\dfrac{1}{3}x^2+\dfrac{2}{9}x^3-\dfrac{14}{81}x^4$
Work Step by Step
Apply Binomial series formula to find the first four terms.
$(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$
and $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$
Thus, we have
$\dfrac{x}{\sqrt[3]{1+x}}=x (1+x)^{-1/3}=x[1+\dfrac{-1}{3}x+\dfrac{(\dfrac{-1}{3})(-\dfrac{4}{3})x^2}{2!}+\dfrac{(\dfrac{-1}{3})(-\dfrac{4}{3})(-\dfrac{7}{3})x^3}{3!}+...]=x-\dfrac{1}{3}x^2+\dfrac{2}{9}x^3-\dfrac{14}{81}x^4+..$
Hence, we have the first four terms:: $x-\dfrac{1}{3}x^2+\dfrac{2}{9}x^3-\dfrac{14}{81}x^4$
