If $$a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}$$   and $$b{\text{ = }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}$$   then the value of $$\left( {\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}}} \right)$$    is = ?

Solution(By Examveda Team)

$$\eqalign{ & a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2} + {{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left[ {{{\left( {\sqrt 5 } \right)}^2} + 1} \right]}}{{5 - 1}} \cr & = \frac{{2\left( {5 + 1} \right)}}{4} \cr & = 3 \cr & a.b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = 1 \cr & {\text{Put value in expression}} \cr & \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & = \frac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a + b} \right)}^2} - 3ab}} \cr & = \frac{{{3^2} - 1}}{{{3^2} - 3}} \cr & = \frac{{9 - 1}}{{9 - 3}} \cr & = \frac{4}{3} \cr} $$