<s>[INST] 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.
 [/INST] 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, ...
</s>