<s>[INST] Is there a quick way to arrive to this partial fraction decomposition? I'm reading a book where the author claims without showing the work that after we go through with the algebra of partial fractions we arrive to the formula (note $|z|<1$):
$$\frac{1}{(1-z)(1-z^2)(1-z^3)} = \frac{1}{6}\frac{1}{(1-z)^3}+\frac{1}{4}\frac{1}{(1-z)^2}+\frac{1}{4}\frac{1}{(1-z^2)}+\frac{1}{3}\frac{1}{(1-z^3)}$$
Naturally, I'm trying to reproduce the result, but I'm taking a very naive approach, namely I've expressed the LHS as a product of irreducible factors over $\mathbb{R}$ and I am trying to determine the coefficients:
$$\frac{1}{(1-z)^3}\frac{1}{(1+z)}\frac{1}{(z^2+z+1)} = \frac{A}{(1-z)}+\frac{B}{(1-z)^2}+\frac{C}{(1-z)^3}+\frac{D}{(1+z)}+\frac{Ex+F}{(z^2+z+1)}$$
This seems to have some drawbacks though, because on top of being long winded, once I do obtain all the coefficients I will have to recombine some of the terms to arrive at the author's answer. My question is then: is there some sort of trick which we can use here which I am not aware of, or do we have to suffer through the algebra patiently?
 [/INST] Here's a bit of a shortcut (if you already know the coefficients). 
Multiply the whole thing by $(1-z)(1-z^2)(1-z^3)$, to get 
\begin{align}
& \frac{1}{6}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^3}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z)^2}+\frac{1}{4}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^2)}+\frac{1}{3}\frac{(1-z)(1-z^2)(1-z^3)}{(1-z^3)}\\ \ \\
&= \frac{(1+z)(1+z+z^2)}{6}+\frac{(1+z)(1-z^3)}{4}+\frac{(1-z)(1-z^3)}{4}+\frac{(1-z)(1-z^2)}{3}\\ \ \\
&= \frac{2(1+z)(1+z+z^2)+3(1+z)(1-z^3)+3(1-z)(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\
&=\frac{2(1+z)(1+z+z^2)+6(1-z^3)+4(1-z)(1-z^2)}{12}\\ \ \\
&=\frac{2z^3+4z^2+4z+2+6-6z^3+4-4z-4z^2+4z^3}{12}\\ \ \\
&=\frac{12}{12}=1.
\end{align}
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