Let α ,   β be the roots of the quadratic equation x 2 + 6 x + 3 = 0 . Then α 23 + &#946…

To find the value of $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$,

Let $a_n={\alpha }^n+{\beta }^n$

Hence,

$\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}=\frac{a_{23}+a_{14}}{a_{15}+a_{10}}$

Now, $x^2+\sqrt{6}x+3=0$ has roots $\alpha \& \beta$

So, $x=\frac{-\sqrt{6}\pm \sqrt{-6}}{2}$

$\Rightarrow x=\sqrt{6}\left(\frac{-1\pm i}{2}\right)$

$\Rightarrow x=\sqrt{3}\left(\frac{-1\pm i}{\sqrt{2}}\right)$

Hence, $\alpha =\sqrt{3} e^{\frac{i3\pi }{4}}$ and $\beta =\sqrt{3}{e}^{\frac{i5\pi }{4}}$

Now, solving $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$

$=\left[\frac{{\left(\sqrt{3}\right)}^{23}\left(e^{i23\times {\displaystyle \frac{3\pi }{4}}}+e^{i23\times {\displaystyle \frac{5\pi }{4}}}\right)+{\left(\sqrt{3}\right)}^{14}\left(e^{i14\times {\displaystyle \frac{3\pi }{4}}}+e^{i14\times {\displaystyle \frac{5\pi }{4}}}\right)}{{\left(\sqrt{3}\right)}^{15}\left(e^{i15\times {\displaystyle \frac{3\pi }{4}}}+e^{i15\times {\displaystyle \frac{5\pi }{4}}}\right)+{\left(\sqrt{3}\right)}^{10}\left(e^{i10\times {\displaystyle \frac{3\pi }{4}}}+e^{i\times 10\times {\displaystyle \frac{5\pi }{4}}}\right)}\right]$

$={\left(\sqrt{3}\right)}^4\left[\frac{{\left(\sqrt{3}\right)}^9\left(e^{i23\times {\displaystyle \frac{3\pi }{4}}}+e^{i23\times {\displaystyle \frac{5\pi }{4}}}\right)+\left(e^{i14\times {\displaystyle \frac{3\pi }{4}}}+e^{i14\times {\displaystyle \frac{5\pi }{4}}}\right)}{{\left(\sqrt{3}\right)}^5\left(e^{i15\times {\displaystyle \frac{3\pi }{4}}}+e^{i15\times {\displaystyle \frac{5\pi }{4}}}\right)+\left(e^{i10\times {\displaystyle \frac{3\pi }{4}}}+e^{i\times 10\times {\displaystyle \frac{5\pi }{4}}}\right)}\right]$

$=9\times \left(\frac{{\left(\sqrt{3}\right)}^9\left({\displaystyle \frac{1+i-1+i}{\sqrt{2}}}\right)+0}{{\left(\sqrt{3}\right)}^5\left({\displaystyle \frac{1+i-1+i}{\sqrt{2}}}\right)+0}\right)$$\left\{\text{as} e^{i\frac{21\mathrm{\pi }}{2}}+e^{i\frac{35\mathrm{\pi }}{2}}=0+i\sin \frac{21\mathrm{\pi }}{2}+0+i\sin \frac{35\mathrm{\pi }}{2}=i-i=0\right\}$

$=9\times \left(\frac{{\left(\sqrt{3}\right)}^9\left({\displaystyle \frac{2i}{\sqrt{2}}}\right)}{{\left(\sqrt{3}\right)}^5\left({\displaystyle \frac{2i}{\sqrt{2}}}\right)}\right)$

$=9\times {\left(\sqrt{3}\right)}^4$

$=81$