If $x-\frac{1}{x}=\sqrt{77}$, then one of the values of $x^3+\frac{1}{x^3}$ is :
Answer & explanation
Correct answer: option 2
If x - \(\frac{1}{x}\) = n
then
then, x + \(\frac{1}{x}\) = \(\sqrt {n^2 + 4}\)
If $x-\frac{1}{x}=\sqrt{77}$
x + \(\frac{1}{x}\) = \(\sqrt {77 + 4}\) = 9
If x + \(\frac{1}{x}\) = n
then, $x^3 +\frac{1}{x^3}$ = n3 - 3 × n
$x^3 +\frac{1}{x^3}$ = 93 - 3 × 9 = (+-) 702