Here is my question. Find the power series, centered at -3 substitute $u = z+3$ and starting at $n = 0$ for the function given by $g(z) = \frac{1}{(1 + 4z) ^ 3}$.

So, this is what I know:
I must differentiate or integrate it or substitute it. I need to use a known series like the following:
$$\sum\_{n=0}^{\infty}x^n=\frac{1}{(1-x)}, x\in]-1,1[$$
or $$\sum\_{n=0}^{\infty}(-1)^nx^2n=\frac{1}{(1-x^2)}, x\in]-1,1[$$  

or $$\sum\_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=\sin(x), x\in \mathbb{R}$$
or $$\sum\_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)}=\arctan(x), x\in [-1,1]$$
or $$\sum\_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}=\cos(x), x\in \mathbb{R}$$

I am not sure how to solve it. Suggestions will be welcome. Thank you.