$\begin{aligned} \therefore \quad \alpha & =\frac{-1+i \sqrt{3}}{2} \text { and } \beta=\frac{-1-i \sqrt{3}}{2} \\ \therefore \quad \alpha \beta & =\left(\frac{-1+i \sqrt{3}}{2}\right)\left(\frac{-1-i \sqrt{3}}{2}\right) \\ & =\frac{(-1)^2-(i \sqrt{3})^2}{4} \\ & =\frac{1-(-1)(3)}{4} \\ & =\frac{1+3}{4}\end{aligned}$
$\therefore \quad \alpha \beta=1$
Also, $\begin{aligned} \alpha+\beta & =\frac{-1+i \sqrt{3}}{2}+\frac{-1-i \sqrt{3}}{2} \\ & =\frac{-1+i \sqrt{3}-1-i \sqrt{3}}{2}=\frac{-2}{2}\end{aligned}$
$\begin{aligned} \therefore \quad & \alpha+\beta=-1 \\ \therefore \quad & (1-\alpha)(1-\beta)\left(1-\alpha^2\right)\left(1-\beta^2\right) \\ & =(1-\alpha)(1-\beta)(1-\alpha)(1+\alpha)(1-\beta)(1+\beta) \\ & =(1-\alpha)^2(1-\beta)^2(1+\alpha)(1+\beta) \\ & =[(1-\alpha)(1-\beta)]^2(1+\alpha)(1+\beta) \\ & =(1-\beta-\alpha+\alpha \beta)^2(1+\alpha+\beta+\alpha \beta) \\ & =[1-(\alpha+\beta)+\alpha \beta]^2[1+(\alpha+\beta)+\alpha \beta] \\ & =[1-(-1)+1]^2(1-1+1) \\ & =3^2(1)=9\end{aligned}$