Group Them Carefully 5

There are two ways to approach this, but I will let someone else write the basics. Notice that $\displaystyle x+\frac{1}{x} = 2\cos(\frac{\pi}{6})\Rightarrow x =\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})$ Hence $x^8 = \cos(\frac{4\pi}{3})+i\sin(\frac{4\pi}{3});\:\frac{1}{x^8} = \cos(\frac{4\pi}{3})-i\sin(\frac{4\pi}{3})$ And $x^8+\frac{1}{x^8} = 2\cos(\frac{4\pi}{3}) = 2(-\frac{1}{2}) = -1$

Guess I'll write the simple solution.

Let's follow the hint and square the equation again:

$\left(x^2 + \dfrac{1}{x^2}\right)^2 = 1^2$

$x^4 + 2 + \dfrac{1}{x^4} = 1$

$x^4 + \dfrac{1}{x^4} = -1$

After that, square it again!

$\left(x^4 + \dfrac{1}{x^4}\right)^2 = (-1)^2$

$x^8 + 2 + \dfrac{1}{x^8} = 1$

Therefore,

$x^8 + \dfrac{1}{x^8} = \boxed{-1}$

Is that x imaginary number?