Given that $x^8-34 x^4+1=0, x > 0$. What is the value of $\left(x^3-x^{-3}\right)$ ?

14

If x⁴ + \(\frac{1}{x^4}\) = a

then x² + \(\frac{1}{x^2}\) = \(\sqrt {a + 2}\) = b

and x - \(\frac{1}{x}\) = \(\sqrt {b - 2}\)

$x^8-34 x^4+1=0, x > 0$

What is the value of $\left(x^3-x^{-3}\right)$ = ?

We can write $x^8-34 x^4+1=0, x > 0$ as x⁴ + \(\frac{1}{x^4}\) = 34

x⁴ + \(\frac{1}{x^4}\) = 34

then x² + \(\frac{1}{x^2}\) = \(\sqrt {34 + 2}\) = 6

and x - \(\frac{1}{x}\) = \(\sqrt {6 - 2}\) = 2

$\left(x^3-x^{-3}\right)$ = 2³ + 3 × 2 = 14