If $x^4+\frac{1}{x^4}=727, x>1$, then what is the value of $\left(x-\frac{1}{x}\right)$ ?
Answer & explanation
Correct answer: option 4
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