Answer
$\Sigma_{n=0}^\infty (-1)^n x^{(3n)}$ for $|x| \lt 1$
Work Step by Step
Consider the Taylor Series for $\dfrac{1}{1-x}$ as follows:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Then, $\dfrac{1}{1+x^3}=\dfrac{1}{1-(x^3)}$
$\implies 1+(-x^3)+(-x^3)^2+....+(-x^3)^n=\Sigma_{n=0}^\infty (-1)^n x^{3n} $
As we can see that $|-x^3| \lt 1$ and $|x| \lt 1$
Therefore, $\dfrac{1}{1+x^3}=\Sigma_{n=0}^\infty (-1)^n x^{(3n)}$ for $|x| \lt 1$
