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

Explanation:-

$$\eqalign{ & x = \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \times \frac{{\left( {\sqrt 3 + 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 + 1} \right)}^2}}}{{\left( {3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{3 + 1 + 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 + \sqrt 3 \cr & y = \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}} \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}} \cr & \,\,\,\,\,\, = \frac{{3 + 1 - 2\sqrt 3 }}{2} \cr & \,\,\,\,\,\, = 2 - \sqrt 3 \cr & \therefore {x^2} + {y^2} = {\left( {2 + \sqrt 3 } \right)^2} + {\left( {2 - \sqrt 3 } \right)^2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\left( {4 + 3} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr}$$