Solve the Following Questions


If $\alpha$ and $\beta$ are the distinct roots of the equation $x^{2}+(3)^{1 / 4} x+3^{1 / 2}=0$, then the value of $\alpha^{96}\left(\alpha^{12}-1\right)+\beta^{96}\left(\beta^{12}-1\right)$ is equal to :

  1. $56 \times 3^{25}$

  2. $56 \times 3^{24}$

  3. $52 \times 3^{24}$

  4. $28 \times 3^{25}$

Correct Option: , 3


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

$\Rightarrow\left(\alpha^{4}+2 \sqrt{3} \alpha^{2}+3\right)=\sqrt{3} \alpha^{2}$ (On squaring)

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

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

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

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

(Multiply by $\alpha^{4}$ )

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

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

Hence, $\alpha^{12}=(27)^{2}$

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