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

Solution (By Examveda Team)

$$\eqalign{ & x = 3 + 2\sqrt 2 {\text{ and }}xy = 1 \cr & {y^2} = \frac{1}{{{x^2}}} \cr & y = \frac{1}{x} = \frac{1}{{3 + 2\sqrt 2 }} = 3 - 2\sqrt 2 \cr & \therefore x + \frac{1}{x} = 3 + 2\sqrt 2 + 3 - 2\sqrt 2 = 6 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 36 - 2 = 34 \cr & \frac{{{x^2} + 3xy + {y^2}}}{{{x^2} - 3xy + {y^2}}} \cr & = \frac{{{x^2} + \frac{1}{{{x^2}}} + 3}}{{{x^2} + \frac{1}{{{x^2}}} - 3}} \cr & = \frac{{34 + 3}}{{34 - 3}} \cr & = \frac{{37}}{{31}} \cr} $$