Answer
$1-\dfrac{1}{2}x^3+\dfrac{3}{8}x^6-\dfrac{5}{16}x^9$
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
$(1+x^3)^{-1/2}=1-\dfrac{1}{2}(x^3)+\dfrac{(\dfrac{-1}{2})(-\dfrac{3}{2})(x^3)}{2!}+\dfrac{(-\dfrac{1}{2})(-\dfrac{3}{2})(-\dfrac{5}{2})(x^3)}{3!}+...=1-\dfrac{1}{2}x^3+\dfrac{3}{8}x^6-\dfrac{5}{16}x^9+...$
Thus, the first four terms are: $1-\dfrac{1}{2}x^3+\dfrac{3}{8}x^6-\dfrac{5}{16}x^9+...$
