Find the power series about the origin for the given function.

(1) $\displaystyle \frac{z^3}{1-z^3}$ , |z| < 1

(2) $\displaystyle z^2cosz$

I will try to attempt the first one:

(1) $\displaystyle \frac{1}{(1-z)^3} = (\frac{1}{1-z})'$

$\displaystyle \frac{1}{1-z} = 1 + z + z^2 + z^3 + ....$

$\displaystyle \frac{z}{1-z} = z + z^2 + z^3 + z^4 + ...$

$\displaystyle \frac{z^3}{1-z} = z^3 + z^6 + z^9 + z^{12} + ...$

$\displaystyle = \sum_{n = 1}^{\infty} z^{3n}$

Somehow, I think my reasoning is wrong. The answer is correct, but I am not sure I took the correct steps.

I have no idea how to approach (2). Thanks for looking and any help provided!