If x - \(\frac{1}{x}\) = 4\(\sqrt{6}\)
then find the value of x3 + \(\frac{1}{x^3}\)
Answer & explanation
Correct answer: option 2
⇒ If x - \(\frac{1}{x}\) = a then x + \(\frac{1}{x}\) = \(\sqrt {a^2 + 4}\)
and
If x + \(\frac{1}{x}\) = a then ⇒ x3 + \(\frac{1}{x^3}\) = a3 - 3a
ATQ,
x - \(\frac{1}{x}\) = 4\(\sqrt{6}\), then
⇒ x + \(\frac{1}{x}\) = \(\sqrt {(4\sqrt{6})^2 + 4}\) = 10
⇒ x3 + \(\frac{1}{x^3}\) = 103 - 3 × 10 = 970