If α and β are the distinct roots of the equation x ^ 2 +...

${x}^{2} + \left(3\right)\dfrac{1}{4}x + 3\dfrac{1}{2} = 0$

Since, $\alpha$ is the root, therefore \left({\alpha}^{2} + \sqrt{3}\right) = - \left(3\right)\dfrac{1}{4} \cdot \alpha$</p><p>On squaring, both sides we get</p><p>\left({\alpha}^{4} + 2\sqrt{3}{\alpha}^{2} + 3\right) = \sqrt{3}{\alpha}^{2}$

$⇒ {\alpha}^{4} + 3 = - \sqrt{3}{\alpha}^{2}$

On squaring, both sides we get

${\alpha}^{8} + 6{\alpha}^{4} + 9 = 3{\alpha}^{4}$

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

Multiply by ${\alpha}^{4}$, we get

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

$∴ {\alpha}^{12} = - 9{\alpha}^{4} - 3 \left(- 9 - 3{\alpha}^{4}\right)</p><p>$⇒ {\alpha}^{12} = 27$</p><p>$⇒ {$\left({\alpha}^{12}\right)$}^{8} = {\left(27\right)}^{8}

$⇒ {\alpha}^{96} = {\left(3\right)}$^{24}$</p><p>Similarly,</p><p>${\beta}^{96} = {\left(3\right)}^{24}

$∴ {\alpha}^{96}\left({\alpha}^{12} - 1\right)$ + {\beta}^{96}$\left({\beta}^{12} - 1\right)=\left(3\right)$24 + \left(27 + 27 - 2\right)$= {\left(3\right)}$^{24} \times 52$<br /> </p>$