If $ x - \frac{1}{x} = 2\sqrt{2}$, then what will be the value of $ x^3 +\frac{1}{x^3}$ ?

$22\sqrt{3}$

If $ x - \frac{1}{x} = 2\sqrt{2}$,

then what will be the value of $ x^3 +\frac{1}{x^3}$

We know that,

If x - \(\frac{1}{x}\)  = n

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

We also know that,

If x + \(\frac{1}{x}\)  = n

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

So, x + \(\frac{1}{x}\)  = \(\sqrt {(2\sqrt{2})^2 + 4}\) = $2\sqrt{3}$

$ x^3 +\frac{1}{x^3}$ = (2$\sqrt{3}$)³ - 3 × 2$\sqrt{3}$ = $22\sqrt{3}$