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If $$x - \frac{1}{x} = 5,$$ x ≠ 0, then what is the value of $$\frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}}?$$
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& {\text{Given, }}x - \frac{1}{x} = 5 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = {5^3} + 3 \times 5 = 140 \cr
& \frac{{{x^6} + 3{x^3} - 1}}{{{x^6} - 8{x^3} - 1}} \cr
& = \frac{{{x^3} + 3 - \frac{1}{{{x^3}}}}}{{{x^3} - 8 - \frac{1}{{{x^3}}}}} \cr
& = \frac{{{x^3} - \frac{1}{{{x^3}}} + 3}}{{{x^3} - \frac{1}{{{x^3}}} - 8}} \cr
& = \frac{{140 + 3}}{{140 - 8}} \cr
& = \frac{{143}}{{132}} \cr
& = \frac{{13}}{{12}} \cr} $$
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