Calculate Inverse Discrete Time Fourier Transform
We need to play a bit with the layout of the expression. We have:

$$
F=\frac{(1+a)(1-a)}{(1-ae^{\ jw})(1-ae^{\ -jw})}
=\frac{(1+a)}{(1-ae^{\ -jw})}\frac{(1-a)}{(1-ae^{\ jw})} 
$$

We can rewrite it as:

$$
=\left( \frac{1}{(1-ae^{\ -jw})}+\frac{a}{(1-ae^{\ -jw})}\right) 
\left( \frac{1}{(1-ae^{\ jw})}-\frac{a}{(1-ae^{\ jw})}\right) \\
$$

We factor out a \$-ae^{\ jw}\$ from the right-most terms and do the inverse transform:

$$
=
\left( \frac{1}{(1-ae^{\ -jw})}+a\frac{1}{(1-ae^{\ -jw})}\right) 
\left( -\frac{1}{a}\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}+\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}\right) \\
\implies \left( a^nu[n]+a(a^nu[n]) \right) * \left( -\tfrac{1}{a}({\tfrac{1}{a}}^nu[n-1])+({\tfrac{1}{a}}^nu[n-1]) \right)
$$

Finally, cleaning up: 

$$
= \left( a^nu[n](1+a) \right) * \left (\tfrac{1}{a}^{n-1}u[n-1](1-\tfrac{1}{a}) \right)
$$

Sorry if it is too messy. Fourier tends to be a lot of writing. Tell me where you think I may have made an error or isn't clear! If someone finds a mistake, please let me know.