Q.If \(a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}\) and \(b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}\), then what is the value of \(\frac{a^2 + ab + b^2}{a^2 - ab + b^2}\)?

\(a + b = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} + \frac{\sqrt{5} - 1}{\sqrt{5} + 1}\) \(= \frac{(\sqrt{5} + 1)^2 + (\sqrt{5} - 1)^2}{(\sqrt{5} - 1)(\sqrt{5} + 1)}\) \(= \frac{2(5 + 1)}{5 - 1}\) \(= \frac{12}{4} = 3\) \(ab = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} \times \frac{\sqrt{5} - 1}{\sqrt{5} + 1} = 1\) Now, \(\frac{a^2 + ab + b^2}{a^2 - ab + b^2}\) \(= \frac{(a + b)^2 - 2ab + ab}{(a + b)^2 - 2ab - ab}\) \(= \frac{(a + b)^2 - ab}{(a + b)^2 - 3ab}\) \(= \frac{3^2 - 1}{3^2 - 3}\) \(= \frac{9 - 1}{9 - 3}\) \(= \frac{8}{6} = \frac{4}{3}\)
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