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

Solution(By Examveda Team)

$$\eqalign{ & {\text{Given,}} \cr & \because a = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}{\text{, }}b = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & {\text{Find }}\frac{{{a^2}}}{b} + \frac{{{b^2}}}{a} = \,? \cr & \Rightarrow \frac{{{a^3} + {b^3}}}{{ab}} = \,? \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - 3ab\left( {a + b} \right)}}{{ab}} = \,? \cr & \Rightarrow a + b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} + \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2} + {{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{{{\sqrt 3 }^2} - {{\sqrt 2 }^2}}} \cr & \Rightarrow \frac{{2\left( {{{\sqrt 3 }^2} + {{\sqrt 2 }^2}} \right)}}{{3 - 2}} \cr & \Rightarrow \frac{{2 \times \left( 5 \right)}}{1} \cr & \Rightarrow a + b = 10 \cr & {\text{Again, }} \cr & \Rightarrow a \times b = \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \Rightarrow ab = 1 \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^3} - 3ab\left( {a + b} \right)}}{{ab}} \cr & \Rightarrow \frac{{{{10}^3} - 3 \times 1 \times 10}}{1} \cr & \Rightarrow 1000 - 30 \cr & \Rightarrow 970 \cr} $$