Answer: Option B
$$\eqalign{
& x - \frac{1}{x} = 1{\text{ }} \cr
& \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr
& {\text{Divide and multiply by }}x \cr
& \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr
& {\text{Take cube on both sides}} \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr
& \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr
& \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr
& \Rightarrow \frac{4}{8} \cr
& \Rightarrow \frac{1}{2} \cr} $$