"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 ${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$ is equal to :"

Question

Accepted Answer

"As, $({\alpha ^2} + \sqrt 3 ) = - {(3)^{1/4}}.\alpha$

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

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

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

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

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

(Multiply by $\alpha$⁴)

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

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

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

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

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

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

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

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

$\Rightarrow$ Option (3) is correct."

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 ${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$ is equal to :

Solution

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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 ${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$ is equal to :

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