Solution:

Given this equation,

$H_a(s)=\frac{s^3}{(s+1)\left(s^2+2 s+2\right)} \text { and } T=1 \mathrm{~s} \text {. }\\ $

To obtain $H(z)$ using the bilinear transformation, put $s=\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)$ in $H_a(s)$.

$H(z)=\left.H_a(s)\right|_{s-\frac{2}{T}\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}=\left.\frac{s^3}{(s+1)\left(s^2+2 s+2\right)}\right|_{s=2\left(\frac{1-z^{-1}}{1+z^{-1}}\right)}\\ $

$\begin{aligned} & =\frac{\left[2 \frac{\left(1-z^{-1}\right)}{\left(1+z^{-1}\right)}\right]^3}{\left[2 \frac{\left(1-z^{-1}\right)}\\{1+z^{-1}}+1\right]\left\{\left[2 \frac{\left(1-z^{-1}\right)}{1+z^{-1}}\right]^2+2\left[2 \frac{\left(1-z^{-1}\right)}\\{1+z^{-1}}\right]+2\right\}} \\\\ & =\frac{8\left(1-z^{-1}\right)^3}{\left[2\left(1-z^{-1}\right)+\left(1+z^{-1}\right)\right]\left[4\left(1-z^{-1}\right)^2+4\left(1-z^{-1}\right)\left(1+z^{-1}\right)+2\left(1+z^{-1}\right)^2\right]}\\ \end{aligned}\\ $

$\begin{aligned}\\ & =\frac{8\left(1-z^{-1}\right)^3}{\left(3-z^{-1}\right)\left[10-4 z^{-1}+2 z^{-2}\right]}\\ \\ & =\frac{4\left(1-z^{-1}\right)^3}{\left(3-z^{-1}\right)\left(5-2 z^{-1}+2 z^{-2}\right)} \\\\ & =4 \frac{\left(1-3 z^{-1}+3 z^{-2}-z^{-3}\right)}{15-11 z^{-1}+8 z^{-2}-2 z^{-3}}\\ \end{aligned}\\ $