Algebra Mess, don't know how to proceed I have kind of an algebra problem. The original question is a Bilinear transformation for analogue to digital filters.
(This is not a homework question)
In my lecture notes, he goes to the answer like 1 step,

I'm trying to work it out but I'm getting stuck at one place and I don't know how to proceed to get the same way he has it.

 A: First, multiply (and divide) by $8(1+z^{-1})^3$: you get 
$$
\frac{8(1+z^{-1})^3}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
$$
Next, expand the numerator to get 
$$
\frac{8(1+3z^{-1}+3z^{-2}+z^{-3})}{8(1+z^{-1})^3+8(1+z^{-1})^2(1-z^{-1})+4(1+z^{-1})(1-z^{-1})^2+(1-z^{-1})^3}.
$$
Now, to expand the denominator. Let's do each term: 
$$
8(1+z^{-1})^3=8+24z^{-1}+24z^{-2}+8z^{-3};
$$
$$
8(1+z^{-1})^2(1-z^{-1})=8+8z^{-1}-8z^{-2}-8z^{-3};
$$
$$
4(1+z^{-1})(1-z^{-1})^2=4-4z^{-1}-4z^{-2}+4z^{-3};
$$
and
$$
(1-z^{-1})^3=1-3z^{-1}+3z^{-2}-z^{-3}.
$$
Collecting terms, the denominator becomes
$$
(8+8+4+1)+(24+8-4-3)z^{-1}+(24-8-4+3)z^{-2}+(8-8+4-1)z^{-3}
=21+25z^{-1}+15z^{-2}+3z^{-3}. 
$$
A: Defining $m:=1-z^{-1}$ and $p:=1+z^{-1}$, we have
$$\begin{align}
\frac{1}{1+2\cdot\dfrac12\dfrac{m}{p}+2\cdot\dfrac14\dfrac{m^2}{p^2}+\dfrac18\dfrac{m^3}{p^3}}\cdot\frac{8p^3}{8p^3} &= \frac{8p^3}{8p^3+8p^2m+4pm^2+m^3} \\[2pt]
&=\frac{8p^3}{\left(2p+m\right)\left(4p^2+2pm+m^2\right)}
\end{align}$$
From here, expanding the various pieces is straightforward.
A: I would set $t=z^{-1}$, to get for the denominator:
\begin{align}
{}&\phantom{={}}\;1+\frac{1-t}{1+t}+\frac{(1-t)^2}{2(1+t)^2}+\frac{(1-t)^3}{8(1+t)^3}\\
&= \frac{8(1+t)^3+8(1-t)(1+t)^2+4(1-t)^2(1+t)+(1-t)^3}{8(1+t)^3} \\
&= \frac{8(1+t)^3+8(1-t^2)(1+t)+4(1-t)(1-t^2)+(1-t)^3}{8(1+t)^3} \\
&=\dotsm
\end{align}
A: Your problem looks very similar to this one.$$\frac{1}{t^3+2t^2+2t+1} = \frac{1}{t^3+3t^2+3t+1-t^2-t}=\frac{1}{(t+1)^3-t(t+1)}=\frac{1}{(t+1)(t^2+t+1)}$$
Can you take it from here?
