If x = 3 + 2 2 and xy = 1, then the value of x^2 + 3xy + y^2x^2 - 3xy + y^2 is?

Answer: Option D

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