If x = √5 + 1 and y = √5 - 1, then what is the value of $$\frac{{{x^2}}}{{{y^2}}} + \frac{{{y^2}}}{{{x^2}}} + 4\left[ {\frac{x}{y} + \frac{y}{x}} \right] + 6?$$

Solution (By Examveda Team)

$$\eqalign{ & x = \sqrt 5 + 1,\,y = \sqrt 5 - 1 \cr & \frac{{{x^2}}}{{{y^2}}} + \frac{{{y^2}}}{{{x^2}}} + 4\left[ {\frac{x}{y} + \frac{y}{x}} \right] + 6 \cr & = {\left[ {\frac{x}{y} + \frac{y}{x}} \right]^2} - 2 + 4\left[ {\frac{x}{y} + \frac{y}{x}} \right] + 6 \cr & = {\left[ {\frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}} \right]^2} + 4\left[ {\frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}} \right] + 4 \cr & = {\left[ {\frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}}} \right]^2} + 4\left[ {\frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}}} \right] + 4 \cr & = {\left[ {\frac{{12}}{4}} \right]^2} + 4\left[ {\frac{{12}}{4}} \right] + 4 \cr & = {\left( 3 \right)^2} + 4\left( 3 \right) + 4 \cr & = 9 + 12 + 4 \cr & = 25 \cr} $$