If x² - 3x + 1 = 0, then the value of $$\frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}$$     will be?

Solution (By Examveda Team)

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{\text{ }} \cr & = \frac{{{x^6}}}{{{x^3}}} + \frac{{{x^4}}}{{{x^3}}} + \frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}} \cr & = {x^3} + \frac{1}{{{x^3}}} + \frac{1}{x} + x \cr & = 18 + 3 \cr & = 21 \cr} $$