If x² - 3x + 1 = 0, then the value of $$\frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}}$$   is:

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & {x^2} + 1 = 3x \cr & x + \frac{1}{x} = 3 \cr & \Rightarrow \frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}} \cr & = \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{\left( {{x^2} + 1 + 5x} \right)}} \cr & = \frac{{x\left[ {{3^3} - 3 \times 3} \right]}}{{\left( {3x + 5x} \right)}} \cr & = \frac{{x\left[ {18} \right]}}{{\left( {8x} \right)}} \cr & = \frac{9}{4} \cr} $$