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

1³/₄
2⁴/₃
3³/₅
4⁵/₃

Answer:- 1

Explanation:-

Solution:

\begin{aligned} a + b = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1} \\ = \frac{{(\sqrt{5} + 1)}^{2} + {(\sqrt{5} - 1)}^{2}}{(\sqrt{5} - 1) (\sqrt{5} + 1)} \\ = \frac{2 [{(\sqrt{5})}^{2} + 1]}{5 - 1} \\ = \frac{2 (5 + 1)}{4} \\ = 3 \\ a . b = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = 1 \\ Put value in expression \\ \frac{a^{2} + a b + b^{2}}{a^{2} - a b + b^{2}} \\ = \frac{{(a + b)}^{2} - a b}{{(a - b)}^{2} - 3 a b} \\ = \frac{3^{2} - 1}{3^{2} - 3} \\ = \frac{9 - 1}{9 - 3} \\ = \frac{4}{3} \end{aligned}

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