I'm not able to show that this expression:

$\displaystyle H_1 = \frac{a^3(1+x^{-1})^3}{(1+a+(a-1)x^{-1})(1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2})}$

can be written

$\displaystyle H_2 = \frac{1}{2} \left( \frac{a-1+(1+a)x^{-1}}{1+a+(a-1)x^{-1}} + \frac{1-a+a^2+(2a^2-2)x^{-1}+(1+a+a^2)x^{-2}}{1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}} \right)$

Partial fraction decomposition of $\displaystyle H_1$ gives

$\displaystyle H_3 = \frac{2(a^3x+a^3-ax+a)}{(a^2+a+1)(a^2x^2+2a^2x+a^2+ax^2-a+x^2-2x+1)} + \frac{a^3}{a^3+2a^2+2a+1} + \frac{2a}{(a+1)(ax+a+x-1)}$

I found $\displaystyle H_3$ by plugging $\displaystyle H_1$ into wolframalpha.

But I can't find a way to $\displaystyle H_2$ from $\displaystyle H_3$ either.

Any hints or things to try?

I have tried to write

$\displaystyle H_1 = \frac{A+Bx^{-1}}{1+a+(a-1)x^{-1}} + \frac{C+Dx^{-1}+Ex^{-2}}{1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}} $

then multiplying through by the denominator gives

$\displaystyle a^3(1+x^{-1})^3 = (A+Bx^{-1})(1+a+a^2+(2a^2-2)x^{-1}+(1-a+a^2)x^{-2}) + (C+Dx^{-1}+Ex^{-2})(1+a+(a-1)x^{-1}) $

and then evaluate at $\displaystyle x^{-1} = -\frac{1+a}{a-1}$ such that it reduces to

$\displaystyle a^3(1-\frac{1+a}{a-1})^3 = (A-B\frac{1+a}{a-1})(1+a+a^2-(2a^2-2)\frac{1+a}{a-1}+(1-a+a^2)(\frac{1+a}{a-1})^2) $

But that gives one equation with 2 unknowns and thus appear to be a dead end!