Overpowered $x$ 's - 2

Note that $\large {(x + \frac {1} {x})} ^ {2} = x^2 + \frac {1} {x^2} +2$ .

If $\large x+ \frac {1} {x} = 1$ then the equation $\large {x} ^ {{2} ^ {n}}+{(\frac {1} {x})} ^ {{2} ^ {n}}= -1$ is valid for all integer $n \geq 1$ .

$\boxed {-1}$

Solving for x yields:

$x + \frac{1}{x} = 1 \Rightarrow x^2 - x + 1 = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 - 4(1)(1)}}{2} = \frac{1 \pm i \sqrt{3}}{2} = e^{\pm i \frac{\pi}{3}}.$

Substituting these roots into the latter power expression gives:

$e^{i \frac{128\pi}{3}} + e^{-i \frac{128\pi}{3}} = 2cos(\frac{128\pi}{3}) = 2(-\frac{1}{2}) = \boxed{-1}.$