Answer
$$\sum_{n=0}^{\infty}(-1)^{n} \frac{x^{3 n}}{2^{3 n+4}}$$
$|x|\lt2 $
Work Step by Step
Given
$$ f(x)=\frac{1}{16+ 2x^3}$$
Since $$\frac{1}{1-x}=\sum_{n=0}^{\infty} x^{n},\ \ \ \ |x|\lt 1\tag{1}$$
By using (1), we get
\begin{align*}
\frac{1}{16+ 2x^3} &=\frac{1}{16(1+ x^3/8)}\\
&=\frac{1}{16} \frac{1}{1+ x^3/8} \\
&=\frac{1}{16} \sum_{n=0}^{\infty}(-x^3/8)^{n}\\
&=\frac{1}{16} \sum_{n=0}^{\infty}\frac{(-1) ^nx^ {3n}}{8^n} \\
&= \sum_{n=0}^{\infty}(-1)^{n} \frac{x^{3 n}}{2^{3 n+4}}
\end{align*} Such that $ |x/2|\lt 1$, or $|x|\lt2 $
