If $\mathrm{\alpha }, \mathrm{\beta }$ are the roots of the equation ${\mathrm{x}}^2-\left(5+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}}\right)x+3\left(3^{{\left({\log }_{3}5\right)}^{\frac{1}{3}}}-5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}-1\right)=0$ then the equation, whose roots are $\mathrm{\alpha }+\frac{1}{\mathrm{\beta }}\text{ and }\mathrm{\beta }+\frac{1}{\mathrm{\alpha }}$,

Bonus because ‘x’ is missing the correct will be,

$\begin{array}{l}{\mathrm{x}}^2-\left(5+3^{\sqrt{{\log }_{3}5}}-5^{\sqrt{{\log }_{5}3}}\right)\mathrm{x}+3\left.3^{{\left({\log }_{3}5\right)}^{\frac{1}{3}}}-5^{{\left({\log }_{5}3\right)}^{\frac{2}{3}}}-1\right)=0\\ 3^{\sqrt{{\log }_{3}5}}=3^{\sqrt{{\log }_{3}5}\cdot \sqrt{{\log }_{3}5}\cdot \sqrt{{\log }_{5}3}}=3^{{\log }_{3}5\cdot \sqrt{{\log }_{5}3}}={\left(3^{{\log }_{3}5}\right)}^{\sqrt{{\log }_{5}3}}=5^{\sqrt{{\log }_{5}3}}\\ 3^{\sqrt[3]{{\log }_{3}5}}=3^{{\log }_{3}5\cdot \sqrt[3]{{\left({\log }_{5}3\right)}^2}}={\left(3^{{\log }_{3}5}\right)}^{{\left({\log }_{5}3\right)}^{2/3}}\\ =5^{{\left({\log }_{5}3\right)}^{2/3}}\end{array}$

So, equation is x² – 5x – 3 = 0 and roots are $\mathrm{\alpha } \& \mathrm{\beta }$

$\left\{\mathrm{\alpha }+\mathrm{\beta }=5; \mathrm{\alpha \beta }=-3\right\}$

New roots are $\mathrm{\alpha }+\frac{1}{\mathrm{\beta }} \& \mathrm{\beta }+\frac{1}{\mathrm{\alpha }}$

$\mathrm{i}.\mathrm{e}., \frac{\mathrm{\alpha \beta }+1}{\mathrm{\beta }} \mathrm{\&} \frac{\mathrm{\alpha \beta }+1}{\mathrm{\alpha }} \mathrm{i}.\mathrm{e}., \frac{-2}{\mathrm{\beta }} \& \frac{-2}{\mathrm{\alpha }}$

Let $\frac{-2}{\mathrm{\alpha }}=\mathrm{t} \Rightarrow \mathrm{\alpha }=\frac{-2}{\mathrm{t}}$

As ${\mathrm{\alpha }}^2-5\mathrm{\alpha }-3=0$

$\begin{array}{l}\Rightarrow {\left(\frac{-2}{\mathrm{t}}\right)}^2-5\left(\frac{-2}{\mathrm{t}}\right)-3=0\\ \Rightarrow \frac{4}{{\mathrm{t}}^2}+\frac{10}{\mathrm{t}}-3=0\\ \Rightarrow 4+10\mathrm{t}-3{\mathrm{t}}^2=0\\ \Rightarrow 3{\mathrm{t}}^2-10\mathrm{t}-4=0\\ \text{ i.e., }3{\mathrm{x}}^2-10\mathrm{x}-4=0\end{array}$