Answer
$\dfrac{1}{1+x^3}=\Sigma_{n=0}^\infty (-1)^n x^{3n}$ for $|x| \lt 1$
Work Step by Step
Since, we know that the Taylor Series for $\dfrac{1}{1-x}$ is defined as:
$\dfrac{1}{1-x}=\Sigma_{n=0}^\infty x^n =1+x+x^2+....+x^n$
Replace $x=-x^3$ in the above form:
$\dfrac{1}{1+x^3}=\dfrac{1}{1-(x^3)}=\Sigma_{n=0}^\infty (-x^3)^n =1+(-x^3)+(-x^3)^2+....+(-x^3)^n=\Sigma_{n=0}^\infty (-1)^n x^{3n} $
Here, $|-x^3| \lt 1 \implies |x| \lt 1$
Hence, $\dfrac{1}{1+x^3}=\Sigma_{n=0}^\infty (-1)^n x^{3n}$ for $|x| \lt 1$
