Algebraic identities

We have, $(x- \frac{1}{x})=2$ On squaring both sides, we get, $(x- \frac{1}{x})^2=4$ $or, x^2 + \frac{1}{x^2} - 2 = 4$ $or, x^2 + \frac{1}{x^2} = 6$ Again squaring both sides, we get, $(x^2 + \frac{1}{x^2})^2 = 36$ $or, x^4 + \frac{1}{x^4} + 2 = 36$ $or, \boxed{x^4 + \frac{1}{x^4} = 34}$

We have $x - \frac{1}{x}=2$ . Let's solve for $x$ :

$x^{2} - 1 = 2x$

$x^{2} - 2x - 1 = 0$

$x = 1 + \sqrt{2}$

We put that value in $x^{4} + \frac{1}{x^{4}}$ :

$(1+\sqrt{2})^{4} + \frac{1}{(1+\sqrt{2})^{4}}$

$17+12\sqrt{2} + \frac{1}{17+12\sqrt{2}}$

$17+12\sqrt{2} + 17-12\sqrt{2}$

$\boxed{34}$