Official solution
\[{3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}\]
\[{3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = 0\]
Note : In the given equation 'x' is missing.
So
\[\alpha + \beta + {1 \over \alpha } + {1 \over \beta } = (\alpha + \beta ) + {{\alpha + \beta } \over {\alpha \beta }}\]
\[ = 5 - {5 \over 3} = {{10} \over 3}\]
\[\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = 2 + \alpha \beta + {1 \over {\alpha \beta }} = 2 - 3 - {1 \over 3} = {{ - 4} \over 3}\]
So Equation must be option (B).