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 :

Given quadratic equation: $x^{2}+\sqrt{6} x+3=0$ <br/><br/>We can find the roots using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ <br/><br/>Here, $a = 1$, $b = \sqrt{6}$, and $c = 3$. <br/><br/>Substituting these values into the quadratic formula, we have: <br/><br/>$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$ <br/><br/>Simplifying further : <br/><br/>$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$ <br/><br/>$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$ <br/><br/>Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $i$ : <br/><br/>$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$ <br/><br/>$\Rightarrow$ $x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$ <br/><br/>$\therefore$ $\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$. <br/><br/>The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation: <br/><br/>$$ \begin{aligned} & =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$ <br/><br/>Here, the exponential power of $\sqrt{3}$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $(\sqrt{3})^{8} = 81$ to simplify: <br/><br/>$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\ \end{aligned} $$ <br/><br/>Since the cosine function has a period of $2\pi$, we can reduce the arguments of the cosine function in the numerator and denominator. <br/><br/>We have $69\pi/4 = \pi/4 + 17\pi = \pi/4$, <br/><br/>$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$, <br/><br/>$45\pi/4 = \pi/4 + 11\pi = \pi/4$, and <br/><br/>$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$. <br/><br/>Therefore, the required expression simplifies to : <br/><br/>$$ \begin{aligned} & =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\ & =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\ & = (\sqrt{3})^{15-7} \\\\ & = (\sqrt{3})^{8} \\\\ & = 81. \end{aligned} $$