[UPLIFTJEE]

Let $\alpha, \beta$ be the roots of the quadratic equation $x^{2}+\sqrt{6} x+3=0$. Then $\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$ is equal to
[JEE (Main) 12 April 2023 S I]
(A) 9
(B) 72
(C) 81
(D) 729

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[UPLIFTJEE]

Solution

$\alpha, \beta=\frac{-\sqrt{6} \pm \sqrt{6} \mathrm{i}}{2}=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$ Required expression $=\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)}=(\sqrt{3})^{8}=81$

[UPLIFTJEE]

Answer

C

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