If x² + 1 = 7x, then the value of \(\frac{x^4\;+\;x^{-2}}{x^2\;+\;9x\;+\;1}\) is?

\(\frac{161}{8}\) 

x² + 1 = 7x

x + \(\frac{1}{x}\) = 7

And, \(\frac{x^4\;+\;x^{-2}}{x^2\;+\;9x\;+\;1}\) = \(\frac{x^3\;+\;\frac{1}{x^3}}{x\;+\;\frac{1}{x}\;+\;9}\)

= \(\frac{(x\;+\;\frac{1}{x})^3\;- 3 (x\;+\;\frac{1}{x})}{(x + \frac{1}{x})\;+\;9}\) = \(\frac{(7)^3\;-\;21}{7\;+\;9}\) = \(\frac{322}{16}\) =  \(\frac{161}{8}\)