As, $\left(a^{2}+\sqrt{3}\right)=-(3)^{1 / 4} \cdot \alpha$

$\Rightarrow\left(\alpha^{2}+2 \sqrt{3} \alpha^{2}+3\right)=\sqrt{3} \alpha^{2} \text { (On squaring) }$

$\left.\therefore\left(a^{4}+3\right)=(-) \sqrt{3} \alpha^{2}\right)$

$\Rightarrow \alpha^{8}+6 \alpha^{4}+9=3 \alpha^{2}$ (Again squaring)

$\therefore \alpha^{8}+3 \alpha^{4}+9=0$

$\Rightarrow \alpha^{8}=-9-3 \alpha^{4}$

(Multiply by $\left.\alpha^{4}\right)$

So, $\alpha^{12}=-9 \alpha^{4}-3 \alpha^{8}$

$\therefore \alpha^{12}=-9 \alpha^{4}-3\left(-9-3 \alpha^{4}\right)$

$\Rightarrow \alpha^{12}=-9 a^{4}+27+9 a^{4}$

Hence, $\alpha^{12}=(27)$

$\Rightarrow\left(\alpha^{12}\right)^{18}=(27)^{8}$

$\Rightarrow \alpha^{96}=(3)^{24}$

Similarly $\beta^{96}=(3)^{24}$

$\therefore \alpha^{96}\left(\alpha^{12}-1\right)+\beta^{96}\left(\beta^{12}-1\right)=(3)^{24} \times 52$