If $$a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$$   & $$b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{,}}$$    then the value of $$\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}{\text{ is?}}$$

Solution(By Examveda Team)

$$\eqalign{ & a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{ }} \cr & \therefore a = \frac{1}{b} \cr & a + b = a + \frac{1}{a} \cr & \Rightarrow \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 1 \right)}^2}}} \cr & \Rightarrow \frac{{6 + 2\sqrt 5 + 6 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow \frac{{12}}{4} \cr & \Rightarrow 3 \cr & \therefore \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & \Rightarrow \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} \cr & \Rightarrow 9 - 2 \cr & \Rightarrow 7\left( {ab = 1} \right) \cr & \therefore \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$