Let α ,β be the roots of the quadratic equation x^2+√6x+3=0. Then α ^23+β ^23+α ^14+β ^14 α ^15+β ^15+α ^10+β ^10 is equal to

To find the value of $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$α15+β15+α10+β10α23+β23+α14+β14​,

Let ${a}_{n}={\alpha }^{n}+{\beta }^{n}$an​=αn+βn

Hence,

$$\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}=\frac{{a}_{23}+{a}_{14}}{{a}_{15}+{a}_{10}}$$α15+β15+α10+β10α23+β23+α14+β14​=a15​+a10​a23​+a14​​

Now, ${x}^{2}+\sqrt{6}x+3=0$x2+6​x+3=0 has roots \alpha &\beta

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

$$\Rightarrow x=\sqrt{6}(\frac{-1\pm i}{2})$$⇒x=6​(2−1±i​)

$$\Rightarrow x=\sqrt{3}(\frac{-1\pm i}{\sqrt{2}})$$⇒x=3​(2​−1±i​)

Hence, $\alpha =\sqrt{3}{e}^{\frac{i3\pi }{4}}$α=3​e4i3π​ and $\beta =\sqrt{3}{e}^{\frac{i5\pi }{4}}$β=3​e4i5π​

Now, solving $\frac{{\alpha }^{23}+{\beta }^{23}+{\alpha }^{14}+{\beta }^{14}}{{\alpha }^{15}+{\beta }^{15}+{\alpha }^{10}+{\beta }^{10}}$α15+β15+α10+β10α23+β23+α14+β14​

$$=[\frac{{(\sqrt{3})}^{23}({e}^{i23\times \frac{3\pi }{4}}+{e}^{i23\times \frac{5\pi }{4}})+{(\sqrt{3})}^{14}({e}^{i14\times \frac{3\pi }{4}}+{e}^{i14\times \frac{5\pi }{4}})}{{(\sqrt{3})}^{15}({e}^{i15\times \frac{3\pi }{4}}+{e}^{i15\times \frac{5\pi }{4}})+{(\sqrt{3})}^{10}({e}^{i10\times \frac{3\pi }{4}}+{e}^{i\times 10\times \frac{5\pi }{4}})}]$$=[(3​)15(ei15×43π​+ei15×45π​)+(3​)10(ei10×43π​+ei×10×45π​)(3​)23(ei23×43π​+ei23×45π​)+(3​)14(ei14×43π​+ei14×45π​)​]

$$={(\sqrt{3})}^{4}[\frac{{(\sqrt{3})}^{9}({e}^{i23\times \frac{3\pi }{4}}+{e}^{i23\times \frac{5\pi }{4}})+({e}^{i14\times \frac{3\pi }{4}}+{e}^{i14\times \frac{5\pi }{4}})}{{(\sqrt{3})}^{5}({e}^{i15\times \frac{3\pi }{4}}+{e}^{i15\times \frac{5\pi }{4}})+({e}^{i10\times \frac{3\pi }{4}}+{e}^{i\times 10\times \frac{5\pi }{4}})}]$$=(3​)4[(3​)5(ei15×43π​+ei15×45π​)+(ei10×43π​+ei×10×45π​)(3​)9(ei23×43π​+ei23×45π​)+(ei14×43π​+ei14×45π​)​]

$=9\times (\frac{{(\sqrt{3})}^{9}(\frac{1+i-1+i}{\sqrt{2}})+0}{{(\sqrt{3})}^{5}(\frac{1+i-1+i}{\sqrt{2}})+0})$=9×((3​)5(2​1+i−1+i​)+0(3​)9(2​1+i−1+i​)+0​)${\text{as}{e}^{i\frac{21\pi }{2}}+{e}^{i\frac{35\pi }{2}}=0+i\mathrm{sin}\frac{21\pi }{2}+0+i\mathrm{sin}\frac{35\pi }{2}=i-i=0}$asei221π​+ei235π​=0+isin221π​+0+isin235π​=i−i=0

$$=9\times (\frac{{(\sqrt{3})}^{9}(\frac{2i}{\sqrt{2}})}{{(\sqrt{3})}^{5}(\frac{2i}{\sqrt{2}})})$$=9×((3​)5(2​2i​)(3​)9(2​2i​)​)

$$=9\times {(\sqrt{3})}^{4}$$=9×(3​)4

$$=81$$=81