Answer
$1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6$
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
$(1+x^2)^{-1/3}=1-\dfrac{1}{3}(x^2)+\dfrac{(\dfrac{-1}{3})(-\dfrac{4}{3})(x^2)^2}{2!}+\dfrac{(-\dfrac{1}{3})(-\dfrac{4}{3})(-\dfrac{7}{3})(x^2)^3}{3!}+...=1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6+...$
Hence, we have the first four terms: $1-\dfrac{1}{3}x^3+\dfrac{2}{9}x^4-\dfrac{14}{81}x^6$
