Correct Answer: $\frac{13}{12}$

Solution : Given: $x-\frac{1}{x}=5$
Cubing both sides, we get
$(x-\frac{1}{x})^3=5^3$
⇒ $x^3-\frac{1}{x^3}-3(x\times\frac{1}{x})(x-\frac{1}{x})=125$
⇒ $x^3-\frac{1}{x^3}-3\times5=125$
⇒ $x^3-\frac{1}{x^3}=140$
Thus,$\frac{x^6+3 x^3-1}{x^6-8 x^3-1}$
Taking $x^3$ as common, we get:
= $\frac{x^3+3-\frac{1}{x^3}}{x^3-8-\frac{1}{ x^3}}$
= $\frac{x^3-\frac{1}{x^3}+3}{x^3-\frac{1}{ x^3}-8}$
Putting the values, we get:
= $\frac{140+3}{140-8}$
= $\frac{143}{132}$
= $\frac{13}{12}$
Hence, the correct answer is $\frac{13}{12}$.