If x² + 1 = 3x, then the value of \(\frac{x^4\;+\;x^{-2}}{x^2\;+\;6x\;+\;1}\) is?

2

x² + 1 = 3x

x + \(\frac{1}{x}\) = 3

And, \(\frac{x^4\;+\;x^{-2}}{x^2\;+\;6x\;+\;1}\) = \(\frac{x^3\;+\;\frac{1}{x^3}}{x\;+\;\frac{1}{x}\;+\;6}\)

= \(\frac{(x\;+\;\frac{1}{x})^3\;- 3 (x\;+\;\frac{1}{x})}{(x + \frac{1}{x})\;+\;6}\) = \(\frac{(3)^3\;-\;9}{3\;+\;6}\) = \(\frac{18}{9}\) = 2