$x^{2}-\left(5+3 \sqrt{\sqrt{\log _{3} 5}}-5 \sqrt{\log _{5} 3}\right) x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$
$3 \sqrt{\log _{3} 5}=3^{\sqrt{\log _{3} 5} \cdot \sqrt{\log _{3} 5} \cdot \sqrt{\log _{5} 3}}=3^{\log _{3} 5 \cdot \sqrt{\log _{5} 3}}=\left(3^{\log _{3} 5}\right)^{\sqrt{\log _{5} 3}}=5^{\sqrt{\log _{5} 3}}$
$3^{\sqrt[3]{\log _{3} 5}}=3^{\log _{3} 5 \sqrt[3]{\left.\log _{5} 3\right)^{2}}}=\left(3^{\log _{3} 5}\right)^{\left(\log _{5} 3\right)^{2 / 3}}=5^{\left(\log _{5} 3\right)^{2 / 3}}$
So, equation is $x ^{2}-5 x -3=0$ and roots are $\alpha \& \beta$
$\{\alpha+\beta=5 ; \alpha \beta=-3\}$
New roots are $\alpha+\frac{1}{\beta} \& \beta+\frac{1}{\alpha}$
i.e., $\frac{\alpha \beta+1}{\beta} \& \frac{\alpha \beta+1}{\alpha}$ i.e., $\frac{-2}{\beta} \& \frac{-2}{\alpha}$
Let $\frac{-2}{\alpha}= t \Rightarrow \alpha=\frac{-2}{ t }$
As $\alpha^{2}-5 \alpha-3=0$
$\Rightarrow\left(\frac{-2}{ t }\right)^{2}-5\left(\frac{-2}{ t }\right)-3=0$
$\frac{4}{t^{2}}+\frac{10}{t}-3=0$
$4+10 t-3 t^{2}=0$
$3 t^{2}-10 t-4=0$
i.e., $3 x^{2}-10 x-4=0$
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