Answer
$1+\dfrac{1}{3}x-\dfrac{1}{9}x^2+\dfrac{5}{81}x^3+...$
Work Step by Step
Formula to find the binomial series is:
$(1+x)^m=1+\Sigma_{k=1}^\infty \dbinom{m}{k}x^k$
Here, $\dbinom{m}{k}=\dfrac{m(m-1)(m-2).....(m-k+1)}{k!}$
Now, $(1+x)^{1/3}=1+\dfrac{1}{3}x+\dfrac{(\dfrac{1}{3})(-\dfrac{2}{3})x^2}{2!}+\dfrac{(\dfrac{1}{3})(-\dfrac{2}{3})(-\dfrac{5}{3})x^3}{3!}+...$
Thus, the first four terms are: $1+\dfrac{1}{3}x-\dfrac{1}{9}x^2+\dfrac{5}{81}x^3+...$
