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.

You could use the binomial expansion as noted in the previous answer but just for fun here's an alternative
Note that:
$$ \dfrac{1}{\left( 1+y \right) ^{4}}=-\dfrac{1}{6}\,{\frac {d^{3}}{d{y}^{3}}}  
 \dfrac{1}{ 1+y }  $$
and that the geometric series gives:
$$  \dfrac{1}{ 1+y }=\sum _{n=0}^{\infty } \left( -y \right) ^{n}$$
so by:
$${\frac {d ^{3}}{d{y}^{3}}}  \left( -y \right) ^{
n}  = \left( -1 \right) ^{n}{y}^{n-3} \left( n-2 \right) 
 \left( n-1 \right) n$$
we have:
$$ \dfrac{1}{\left( 1+y \right) ^{4}}=-\dfrac{1}{6}\,\sum _{n=0}^{\infty } \left( -1
 \right) ^{n}{y}^{n-3} \left( n-2 \right)  \left( n-1 \right) n$$
and putting $y=x^3$ gives:
 $$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=-\dfrac{1}{6}\,\sum _{n=0}^{\infty }{x}^{3\,n-9
} \left( -1 \right) ^{n} \left( n-2 \right)  \left( n-1 \right) n$$
then noting that the first 3 terms are zero because of the $n$ factors we can shift the index by letting $n\rightarrow m+3$ to get:
$$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=\dfrac{1}{6}\,\sum _{m=0}^{\infty }{x}^{3m} \left( -1 \right) ^{m} \left( 1+m
 \right)  \left( m+2 \right)  \left( m+3 \right) $$
For comparison the binomial expansion would tell you that:
$$\dfrac{1}{\left( 1+x^3 \right) ^{4}}=\sum _{m=0}^{\infty }{-4\choose m}{x}^
{3\,m}$$
from which it follows, by the uniqueness of Taylor series, that:
$${-4\choose m}=\dfrac{1}{6} \left( -1 \right) ^{m} \left( 1+m \right)  \left( 
m+2 \right)  \left( m+3 \right)=(-1)^m\dfrac{(3+m)!}{3!\,m!}$$ 
