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.
 A: 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:
$${-4\choose m}=\dfrac{1}{6} \left( -1 \right) ^{m} \left( 1+m \right)  \left( 
m+2 \right)  \left( m+3 \right)=(-1)^m\dfrac{(3+m)!}{3!\,m!}$$ 
A: \begin{gather}(1+x^3)^{-4}=(1+t)^\alpha=1+\alpha t+\frac{\alpha(\alpha-1)}{2!}t^2+\dots=\\ =1+\frac{(-4)}{1!}x^3+\frac{(-4)(-4-1)}{2!}x^6+\dots
=\sum_{k=0}^\infty(-1)^{k}\frac{(3+k)!}{3!k!}x^{3k}\end{gather}
A: 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}
A: 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}
