Answer
$1+\dfrac{1}{3}x-\dfrac{1}{9}x^2+\dfrac{5}{81}x^3$
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,$(1+x)^{1/3}=1+(\dfrac{1}{3})x+\dfrac{(\dfrac{1}{3})(-\dfrac{2}{3})x^2}{2!}+\dfrac{(\dfrac{1}{3})(-\dfrac{2}{3})(-\dfrac{5}{3})x^3}{3!}+...=1+\dfrac{1}{3}x-\dfrac{1}{9}x^2+\dfrac{5}{81}x^3+...$
Hence, we have the first four terms are: $1+\dfrac{1}{3}x-\dfrac{1}{9}x^2+\dfrac{5}{81}x^3$
