If $$a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }},$$   $$b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}$$   then the value of $${a^2} + {b^2}$$   would be = ?

Solution (By Examveda Team)

$$\eqalign{ & \because a = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 + 2\sqrt 6 }}{{3 - 2}} \cr & = 5 + 2\sqrt 6 \cr & \because b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \cr & {\text{ = }}\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{ = }}\frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{{{\left( {\sqrt 3 } \right)}^2} - {{\left( {\sqrt 2 } \right)}^2}}} \cr & = \frac{{3 + 2 - 2\sqrt 6 }}{{3 - 2}} \cr & = 5 - 2\sqrt 6 \cr & \therefore {\text{ }}{a^2} + {b^2} \cr & = {\left( {5 + 2\sqrt 6 } \right)^2} + {\left( {5 - 2\sqrt 6 } \right)^2} \cr & = 2\left[ {{{\left( 5 \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} \right] \cr & = 2\left( {25 + 24} \right) \cr & = 2 \times 49 \cr & = 98 \cr} $$