If $$x^2-3x-1=0$$, then the value of $$(x^2+8x-1)(x^3+x^{-1})^{-1}$$ is:

Given, $$x^2-3x-1=0$$

$$\Rightarrow$$  $$x\left(x-3-\frac{1}{x}\right)=0$$

$$\Rightarrow$$  $$x-3-\frac{1}{x}=0$$

$$\Rightarrow$$  $$x-\frac{1}{x}=3$$ .........(1)

$$(x^2+8x-1)(x^3+x^{-1})^{-1}=\frac{(x^2+8x-1)}{(x^3+\frac{1}{x})}$$

$$=\frac{x(x+8-\frac{1}{x})}{x(x^2+\frac{1}{x^2})}$$

$$=\frac{(x-\frac{1}{x}+8)}{x^2+\frac{1}{x^2}-2+2}$$

$$=\frac{3+8}{\left(x-\frac{1}{x}\right)^2+2}$$

$$=\frac{11}{\left(3\right)^2+2}$$

$$=\frac{11}{11}$$

$$=1$$

Hence, the correct answer is Option C