Answer
The power series for $f(x)$ is $\sum_{n=0}^\infty \frac{x^{3n+2}}{a^{3n+3}}$
The interval of convergence is $(-|a|,|a|)$
Work Step by Step
$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$ where $|x|<1$
Replacing $x$ by $x^3/a^3$,
$\frac{1}{1-x^3/a^3}=\sum_{n=0}^\infty (x^3/a^3)^n$ where $|x^3/a^3|<1$
$\frac{a^3}{a^3-x^3}=\sum_{n=0}^\infty \frac{x^{3n}}{a^{3n}}$ where $|x|^3<|a|^3$
Multiplying by $\frac{x^2}{a^3}$,
$\frac{x^2}{a^3-x^3}=\sum_{n=0}^\infty \frac{x^{3n+2}}{a^{3n+3}}$ where $|x|<|a|$
