If \(x^4 + \frac{1}{x^4} = 119\) and \(x > 1\), what is the value of \(x^3 - \frac{1}{x^3}\)?

A. 33

B. 36

C. 39

D. 27

Doubtly AI Asked 27 Oct 2026 AI-generated · not yet verified ssc cgl Save WhatsApp Report

1 answer

Best answer

Answer: B. 36

Answer: (B) 36

\(\left(x^2 + \frac{1}{x^2}\right)^2 = x^4 + \frac{1}{x^4} + 2 = 121\), so \(x^2 + \frac{1}{x^2} = 11\).

\(\left(x - \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} - 2 = 9\). Since \(x > 1\), \(x - \frac{1}{x} = 3\).

Shortcut: \(x^3 - \frac{1}{x^3} = k^3 + 3k\) where \(k = x - \frac{1}{x} = 3\): 27 + 9 = 36. (Here \(x = \frac{3 + \sqrt{13}}{2}\), a real number.)

Doubtly AI · 27 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?