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If \( \Large \left( x^{2} + \frac{1}{x^{2}} \right) = \frac{17}{4}\), then what is \( \Large \left( x^{3} - \frac{1}{x^{3}} \right) \) equal to?

A) \( \Large \frac{75}{11} \)
B) \( \Large \frac{63}{8} \)
C) \( \Large \frac{95}{8} \)
D) None of these

Correct Answer:


B) \( \Large \frac{63}{8} \)

Description for Correct answer:
Here \( \Large x^{2} + \frac{1}{x^{2}} = \frac{17}{4} \)

Now \( \Large \left(x - \frac{1}{x}\right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{17}{2} - 2 \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{17 - 8}{4} \)

\( \Large \left(x - \frac{1}{x}\right)^{2} = \frac{9}{4} \)

\( \Large \left(x - \frac{1}{x}\right) = \frac{3}{2} \)

Now

\( \Large \left(x - \frac{1}{x}\right)^{2} = \left(x - \frac{1}{x}\right) \left(x^{2} + 1 + \frac{1}{x^{2}} \right) \)

\( \Large = \left(\frac{3}{2}\right) \left(\frac{17}{4} + 1\right) = \left(\frac{3}{2}\right) \left(\frac{21}{4}\right) \)

\( \Large =\frac{63}{8}\)

Part of solved CDS Maths(2) questions and answers : Exams >> CDSE >> CDS Maths(2)

Comments

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