If $ x - \frac{1}{x} = 2\sqrt{2}$, then what will be the value of $ x^3 +\frac{1}{x^3}$ ?
Answer & explanation
Correct answer: option 1
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$\sqrt{3}$ = $22\sqrt{3}$