If x² - 3x - 1 = 0, then the value of (x² + 8x - 1)(x³ + x^-1)^-1 is:

$$\eqalign{ & {x^2} - 3x - 1 = 0 \cr & x\left( {x - 3 - \frac{1}{x}} \right) = 0 \cr & x - \frac{1}{x} = 3 \cr & {x^2} + \frac{1}{{{x^2}}} = 11 \cr & \frac{{\left( {{x^2} + 8x - 1} \right)}}{{{x^3} + \frac{1}{x}}} \cr & = \frac{{x\left( {\frac{{x - 1}}{{x + 8}}} \right)}}{{x\left( {\frac{{{x^2} + 1}}{{{x^2}}}} \right)}} \cr & = \frac{{3 + 8}}{{11}} \cr & = 1 \cr} $$