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.

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You can also use : $\forall x$ such that $\vert x \vert < 1$, $\frac{1}{1+x} = \sum_{n=0}^{+\infty} (-1)^{n}x^{n}$. –  jibounet Jul 24 '13 at 7:28
Why not expand $(\frac1{1+y})^4$ and then put $y=x^3$? –  Mark Bennet Jul 24 '13 at 7:33
@jibounet Sorry but how (use this hint)? –  Did Jul 24 '13 at 8:25
Why not use the expansion $(1+t)^\alpha=1+\alpha t+\dots$? –  Yurii Savchuk Jul 24 '13 at 9:39
Thank you all. @MarkBennet Thank you. I then get $\binom{3}{0}-\binom{4}{1}y+\binom{5}{2}y^2+\dots+\binom{k+n}{n}y^n = \binom{3}{0}-\binom{4}{1}x^3+\binom{5}{2}x^6+\dots+\binom{k+n}{n}x^{3n}$ upon setting $y=x^3$. –  tychicus Jul 25 '13 at 12:24

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: $${-4\choose m}=\dfrac{1}{6} \left( -1 \right) ^{m} \left( 1+m \right) \left( m+2 \right) \left( m+3 \right)=(-1)^m\dfrac{(3+m)!}{3!\,m!}$$

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Thank you for a thorough answer! –  tychicus Jul 25 '13 at 12:26

\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}

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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}

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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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Aaaaargh...  –  Did Jul 25 '13 at 7:04