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.

Just use the generalized binomial series. For natural $m$:
$$
(1 + u)^{-m}
  = \sum_{k \ge 0} \binom{-m}{k} u^k
  = \sum_{k \ge 0} (-1)^k \binom{k + m - 1}{m - 1} u^k
$$
and plug in $u = x^3$, $m = 4$:
\begin{align}
(1 + x^3)^{-4}
  &= \sum_{k \ge 0} (-1)^k \binom{k + 3}{3} x^{3 k} \\
  &= \sum_{k \ge 0} (-1)^k \frac{(k + 3) (k + 2) (k + 1)}{3!} x^{3 k} \\
  &= \frac{1}{6} \sum_{k \ge 0} (-1)^k (k^3+ 6 k^2 + 11 k + 6) x^{3 k}
\end{align}
