If $x^4-62 x^2+1=0$, where $x>0$, then the value of $x^3+x^{-3}$ is:

488

If $x^4-62 x^2+1=0$, where $x>0$,

then the value of $x^3+x^{-3}$

 If, x² + \(\frac{1}{x^2}\) = b

then, x + \(\frac{1}{x}\) = \(\sqrt {b + 2}\)

If $x^4-62 x^2+1=0$

Then, x² + \(\frac{1}{x^2}\) = 62

and, x + \(\frac{1}{x}\) = \(\sqrt {62 + 2}\) = 8

$x^3+x^{-3}$ = 8³ - 3 × 8 = 488