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If $$x - \frac{3}{x} = 6,\,x \ne 0,$$ then the value of $$\frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}}$$
Answer & Solution
Correct Answer:
Option
D
$$\eqalign{
& \frac{{{x^4} - \frac{{27}}{{{x^2}}}}}{{{x^2} - 3x - 3}} \cr
& = \frac{{x\left( {{x^3} - \frac{{27}}{{{x^3}}}} \right)}}{{x\left( {x - 3 - \frac{3}{x}} \right)}}.....\left( {\text{i}} \right) \cr
& {x^3} - \frac{{27}}{{{x^3}}} \cr
& = {6^3} + 3 \times 3 \times 6 \cr
& = 216 + 54 \cr
& = 270 \cr
& \frac{{{x^3} - \frac{{27}}{{{x^3}}}}}{{x - \frac{3}{x} - 3}} = \frac{{270}}{{6 - 3}} = 90 \cr} $$
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