If $$x - \frac{1}{x} = 5,$$   x ≠ 0, then what is the value of $$\frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}}?$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{Given, }}x - \frac{1}{x} = 5 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = {5^3} + 3 \times 5 = 140 \cr & \frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}} \cr & = \frac{{{x^3} + 3 - \frac{1}{{{x^3}}}}}{{{x^3} - 8 - \frac{1}{{{x^3}}}}} \cr & = \frac{{{x^3} - \frac{1}{{{x^3}}} + 3}}{{{x^3} - \frac{1}{{{x^3}}} - 8}} \cr & = \frac{{140 + 3}}{{140 - 8}} \cr & = \frac{{143}}{{132}} \cr & = \frac{{13}}{{12}} \cr} $$