If $$x$$ = $$\sqrt 3 - \frac{1}{{\sqrt 3 }}$$   and $$y$$ = $$\sqrt 3 + \frac{1}{{\sqrt 3 }}$$   then the value of $$\frac{{{x^2}}}{y} + \frac{{{y^2}}}{x}$$  is?

Solution(By Examveda Team)

$$\eqalign{ & x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr & = \frac{{{x^3} + {y^3}}}{{xy}} \cr & = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr & \therefore x + y \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 2\sqrt 3 \cr & \therefore xy \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 3 - \frac{1}{3} \cr & = \frac{8}{3} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr & \Rightarrow 3\sqrt 3 \cr} $$