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