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