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

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, ...
