Inverse $\mathcal Z$-transform of rational functions
Given the $\mathcal Z$-transform :
$$H(z) = \frac{\left(1+\beta z^{-1}\right)\left(1+\beta z\right)}{\left(1+\alpha z^{-1}\right)\left(1+\alpha z\right)}$$

The inverse $\mathcal Z$-transform can be found by the method of partial fraction expansion guided with the Region of Convergence ROC associated with $H(z)$

I will proceed with the most fundamental method and apply necesary algebraic manipulations to get the required standard form as follows:

\begin{align}
H(z) &amp;= \frac {\left(1 + \beta z^{-1}\right)\left(1 + \beta z\right)}{\left(1 + \alpha z^{-1}\right)\left(1 + \alpha z\right)} &amp;\scriptstyle{\text{convert to negative power of $z$}}\\ 
 &amp;= \frac{\left(1 + \beta z^{-1}\right)\left(z^{-1} + \beta\right)}{\left(1 + \alpha z^{-1}\right)\left(z^{-1} + \alpha\right)} &amp;\scriptstyle{\text{convert to standard form of } (1-d_k z^{-1})}\\
&amp;= \frac{\beta}{\alpha} \frac{\left(1 + \beta z^{-1}\right)\left(1 + \frac{1}{\beta} z^{-1}\right)}{\left(1 + \alpha z^{-1}\right)\left(1 + \frac{1}{\alpha}z^{-1}\right)} &amp;\scriptstyle{\text{expand the brackets}}\\
 &amp;= \frac{\beta}{\alpha} \frac{\left(1 + \left(\beta + \frac{1}{\beta}\right) z^{-1} + z^{-2}\right) }{\left(1 + \left(\alpha + \frac{1}{\alpha}\right) z^{-1} + z^{-2}\right)} &amp;\scriptstyle{\text{apply the long division}}\\
&amp;= \frac{\beta}{\alpha} \left( 1 + \frac{ \left(\beta + \frac{1}{\beta} - \alpha - \frac{1}{\alpha}\right)z^{-1} }{\left(1 + \alpha z^{-1}\right)\left(1 + \frac{1}{\alpha}z^{-1}\right)} \right) &amp;\scriptstyle{\text{go further...}}\\
&amp;= \frac{\beta}{\alpha} \left( 1 + \frac{A}{1 + \alpha z^{-1}} + \frac{B}{1 + \frac{1}{\alpha}z^{-1}} \right) &amp;\scriptstyle{\text{where $A$ and $B$ are found from the right side quotient of $H(z)$}}\\
\end{align}

Now assuming $A$ and $B$ are found as functions of $\alpha$ and $\beta$, we shall define all possible ROC of $H(z)$ to find the corresponding inverse transform $h[n]$.


Now If $0&lt;|\alpha|&lt;1&lt;\frac{1}{|\alpha|}$ and ROC is such that $|z|&lt;|\alpha|$ then: $h[n]$ will be unstable and anti-causal as
$$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left(- A\left(-\alpha\right)^n \text{u}[-n-1] - B\left(- \frac{1}{\alpha}\right)^n \text{u}[-n-1] \right)$$ 
Or else if $0&lt;|\alpha|&lt;1&lt;\frac{1}{|\alpha|}$ and ROC is such that $|\alpha| &lt;|z|&lt;\frac{1}{|\alpha|}$ then: $h[n]$ will be stable and two-sided as
$$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left( A(-\alpha)^n \text{u}[n] - B\left(- \frac{1}{\alpha}\right)^n \text{u}[-n-1] \right)$$ 
Or else if $0&lt;|\alpha|&lt;1&lt;\frac{1}{|\alpha|}$ and ROC is such that $|z|&gt;\frac{1}{|\alpha|}$ then: $h[n]$ will be unstable and causal as
$$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left( A(-\alpha)^n \text{u}[n] + B\left(- \frac{1}{\alpha}\right)^n \text{u}[n] \right)$$ 
The cases for $0&lt;\frac 1\alpha&lt;1&lt;\alpha$ are simply the same, merely interchanging $\alpha$ with $\frac 1\alpha$