If $x +\frac{1}{x}=\sqrt{13}$, then one of the values of $x^3 -\frac{1}{x^3}$ is :

36

If $x +\frac{1}{x}=\sqrt{13}$,

then one of the values of $x^3 -\frac{1}{x^3}$ is = ?

We know that,

If x + \(\frac{1}{x}\)  = n

then, x - \(\frac{1}{x}\)  = \(\sqrt {n^2 - 4}\)

and we also know that,

If x - \(\frac{1}{x}\)  = n

then, $x^3 -\frac{1}{x^3}$ = n³ + 3 × n

x - \(\frac{1}{x}\) = \(\sqrt {\sqrt{13}^2 - 4}\) = 3

Then, $x^3 -\frac{1}{x^3}$ = 3³ + 3 × 3

$x^3 -\frac{1}{x^3}$ = 27 + 9 = 36