If $x^4+\frac{1}{x^4}=727, x>1$, then what is the value of $\left(x-\frac{1}{x}\right)$ ?

5

If x⁴ + \(\frac{1}{x^4}\) = a

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

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

If $x^4+\frac{1}{x^4}=727, x>1$

then x² + \(\frac{1}{x^2}\) = \(\sqrt {727 + 2}\) = 27

and x - \(\frac{1}{x}\) = \(\sqrt {27 - 2}\) = 5