TITLE: Laurent Series expansion without geometric series
QUESTION [5 upvotes]: There are several functions in complex analysis which I have not been able to get the Laurent expansion for, both of which are very different from the examples I see online and in the (4) textbooks I have checked out...:
I need to find the Laurent expansion about each singularity of the following function: 
$$f(z) = {1 \over z^6+1}$$
I had no issue with finding the singular points, but I don't see how to create a Laurent expansion from there---all of the online examples show something like:
$$f(x) = {1 \over z(z-1)}$$ 
In which it is much more clear how to use a geometric series to find the Laurent series.
I also have the same issue for the following function:
$$f(z) = {1 \over z^4+2z^2+1}$$
I can find the singularities, but where do I go from there? The examples found online are tough to map onto these problems.

REPLY [2 votes]: Let $\alpha=e^{(2n+1)\pi i/6}$ be one of the roots of $z^6+1$ and $\alpha w=z-\alpha$.
$$
\begin{align}
\frac1{z^6+1}
&=\frac1{1-(1+w)^6}\tag{1}\\
&=-\frac1w\frac1{6+15w+20w^2+15w^3+6w^4+w^5}\tag{2}\\
&=\sum_{k=-1}^\infty b_kw^k\tag{3}\\
&=-\frac1{6w}+\frac5{12}-\frac{35}{72}w+\frac{35}{144}w^2+\frac{119}{864}w^3+\dots\tag{4}
\end{align}
$$

Explanation:
  $(1)$: $z^6+1=1+\alpha^6(1+w)^6=1-(1+w)^6$
  $(2)$: Binomial Theorem
  $(3)$: label the powers of $w$ in the expansion of $(2)$
  $(4)$: multiply both sides of $(2)$ and $(3)$ by $w\!\left(6+15w+20w^2+15w^3+6w^4+w^5\right)$:
  $\phantom{(3)\,}$ $\color{#C00000}{-1}=\left(6+15w+20w^2+15w^3+6w^4+w^5\right)\sum\limits_{k=-1}^\infty b_kw^{k+1}$
  $\phantom{(3)\,}$ $\phantom{-1}=\color{#C00000}{6b_{-1}}+\color{#00A000}{(15b_{-1}+6b_0)}w+\color{#00A000}{(20b_{-1}+15b_0+6b_1)}w^2$
  $\phantom{(3)\,}$ $\phantom{-1}+\color{#00A000}{(15b_{-1}+20b_0+15b_1+6b_2)}w^3+\color{#00A000}{(6b_{-1}+15b_0+20b_1+15b_2+6b_3)}w^4$
  $\phantom{(3)\,}$ $\phantom{-1}+\sum\limits_{k=4}^\infty\color{#0000F0}{(b_{k-5}+6b_{k-4}+15b_{k-3}+20b_{k-2}+15b_{k-1}+6b_k)}w^{k+1}$
  $\phantom{(3)\,}$ The red term is $-1$ and gives $b_{-1}=-\frac16$
  $\phantom{(3)\,}$ The green terms are $0$ and give the other coefficients in $(4)$.
  $\phantom{(3)\,}$ The blue term in the sum is $0$ and gives the recursion in $(5)$.

where, for $k\ge4$,
$$
b_k=-\frac{15b_{k-1}+20b_{k-2}+15b_{k-3}+6b_{k-4}+b_{k-5}}6\tag{5}
$$
Then substitute $w=\frac{z-\alpha}\alpha$ into $(3)$ to get
$$
\frac1{z^6+1}=\sum_{k=-1}^\infty\frac{b_k}{\alpha^k}(z-\alpha)^k\tag{6}
$$