How to convert $\frac{1}{(1-x)(1-x^3)}$ into a sum of multiple fractions?

Question

What I mean is, that one can convert $\frac{1}{(1-x)^2(1-x^2)}$ into the following sum:

$\frac{1}{8}(\frac{1}{1+x}+\frac{1}{1-x} +\frac{2}{(1-x)^2} + \frac{4}{(1-x^3)})$

But I can't seem to do the same here, because when I try to simplify $1-x^3$ I get $(1-x)(1+x+x^2)$, and the second term can't be further simplified.

For context, this is part of another problem in combinatorics, which I'm trying to solve using generating functions.

Answer 1

Since you know that$$\frac1{(x-1)(x^3-1)}=\frac1{(x-1)^2(x^2+x+1)},$$use this to deduce that$$\frac1{(x-1)(x^3-1)}=\frac{x+1}{3\left(x^2+x+1\right)}-\frac1{3(x-1)}+\frac1{3 (x-1)^2},$$by partial fraction decomposition.

Answer 2

To get the exact answer requested first use a difference of two squares; $$\frac1{(1-x)^2(1-x^2)}=\frac1{(1+x)(1-x)^3} $$The standard partial fraction treatment for repeated factors gives $$\frac1{(1+x)(1-x)^3}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^2}+\frac{D}{(1-x)^3} $$ $$1=A(1-x)^3+B(1+x)(1-x)^2+C(1+x)(1-x)+D(1+x) $$ $x$ = 1 gives $D$ = $\frac{1}{2}$, $x$ = -1 gives $A$ = $\frac{1}{8}$, matching coefficients of $x^3$ gives $B$ =$\frac{1}{8}$, and $x$ = 0 gives $C$ = $\frac{1}{4}$ \begin{equation} \frac1{(1+x)(1-x)^3}=\frac{1}{8} \bigg\{ \frac{1}{(1+x)}+\frac{1}{(1-x)}+\frac{2}{(1-x)^2}+\frac{4}{(1-x)^3} \bigg\} \end{equation} It generates the sequence 1, 2, 4, 6, 9, 12, 16, 20, 25, 30, 36, 42, 49, 56, 64, 72, ...