<s>[INST] 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.
 [/INST] 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}
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