If and are the distinct roots of x^2+3^1/4x+3^1/2=0, then the value of ^96(^12-1)+^96(^12-…

Question

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

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.

Solutions · 1