Power series for $(1+x^3)^{-4}$ I am trying to find the power series for the sum $(1+x^3)^{-4}$ but I am not sure how to find it. Here is some work: 
$$(1+x^3)^{-4} = \frac{1}{(1+x^3)^{4}} = \left(\frac{1}{1+x^3}\right)^4 = \left(\left(\frac{1}{1+x}\right)\left(\frac{1}{x^2-x+1}\right)\right)^4$$
I can now use 
$$\frac{1}{(1-ax)^{k+1}} = \left(\begin{array}{c} k \\ 0 \end{array}\right)+\left(\begin{array}{c} k+1 \\ 1 \end{array}\right)ax+\left(\begin{array}{c} k+2 \\ 2 \end{array}\right)a^2x^2+\dots$$ 
on the $\frac{1}{1+x}$ part but I am not sure how to cope with the rest of formula.

This would not be the best answer but may help if you are interested in computing the first terms of the Taylor expansion.
Just use the fact that $\frac{1}{1-z}= \sum_{n=0}^\infty z^n $ for $|z|<1$ and the binomial formula, i.e.
\begin{align}
\frac{1}{(1+x^3)^4} &= \frac{1}{1+4x^3+6x^6+4x^9+x^{12}} \\
&= \sum_{n=0}^\infty (-1)^n (4x^3+6x^6+4x^9+x^{12})^n.
\end{align}
