Option 1 : 78

**Given:**

\(x - \frac{3}{x} = 6\)

**Concept used:**

Algebra

**Calculation:**

⇒ \(\frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}}\)

Divide the above fraction by 'x' in both numerator and denominator

⇒ \(\frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - 3 - \frac{3}{x}}}\)

⇒ \(x - \frac{3}{x} = 6\) (Given)

⇒ if \(a - \frac{1}{a} = k\ {\rm{ then\ }}{{\rm{a}}^3} - \frac{1}{{{{\rm{a}}^3}}} = {\rm{ }}{{\rm{k}}^3} + 3k\)

⇒ \({x^3} - \frac{27}{{{x^3}}} = 216 + 3\left( {6} \right)\)

⇒ \({x^3} - \frac{{27}}{{{x^3}}} = 234 \)

∴ 270/3 = **78**