If x = 1 + $$\sqrt 2 $$  + $$\sqrt 3 $$  and y = 1 + $$\sqrt 2 $$  - $$\sqrt 3 {\text{,}}$$  then the value of $$\frac{{{x^2} + 4xy + {y^2}}}{{x + y}}$$   is?

$$\eqalign{ & {\text{According to the question,}} \cr & x = 1 + \sqrt 2 + \sqrt 3 \,.....(i) \cr & y = 1 + \sqrt 2 - \sqrt 3 \,.....(ii) \cr & \Rightarrow \frac{{{x^2} + 4xy + {y^2}}}{{x + y}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & {\text{From equation (i)}} + {\text{(ii)}} \cr & \Rightarrow x + y = 2 + 2\sqrt 2 \cr & xy = {\left( {1 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr & \Rightarrow xy = 3 + 2\sqrt 2 - 3 \cr & \Rightarrow xy = 2\sqrt 2 \cr & {\text{So, }}\frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & = \frac{{{{\left( {2 + 2\sqrt 2 } \right)}^2} + 2 \times 2\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{4 + 8 + 8\sqrt 2 + 4\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12 + 12\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12\left( {1 + \sqrt 2 } \right)}}{{2\left( {1 + \sqrt 2 } \right)}} \cr & = 6 \cr} $$