Solução padrão: $(x+\frac{1}{x})^2 = 1$ $\implies$ $x^2+\frac{1}{x^2} = -1$ $(x^2+\frac{1}{x^2})^2 = (-1)^2 = 1$ $\implies$ $x^4+\frac{1}{x^4} = -1$ $(x^4+\frac{1}{x^4})\cdot (x^2+\frac{1}{x^2})$ $=$ $(x^6+\frac{1}{x^6}) + (x^2+\frac{1}{x^2})$ $=$ $(-1)\cdot (-1)$ $=$ $1$ $\large{(x^6+\frac{1}{x^6})}$ $+$ $\underbrace{\left(x^2+\frac{1}{x^2} \right)}_{-1}$ $=$ $1$ $\implies$ $\boxed{\left(x^6+\frac{1}{x^6} \right) = 2}$ Solução brilhante: Consideremos $x = \pu{cis}(\theta)$. Sabendo que $2 \pu{cos}\left(\dfrac{\pi}{3}\right)~=~1$, temos que:$$x+\dfrac{1}{x}~=~ \pu{cis}(\theta)+ \pu{cis}(-\theta)~=~2 \pu{cos}(\theta)~=~2 \pu{cos}\left(\dfrac{\pi}{3}\right)\implies \theta~=~\dfrac{\pi}{3}$$Dessa forma, $x~=~\pu{cis}\left(\dfrac{\pi}{3}\right)$ e então, a partir da Lei de Moivre:$$x^6 + \dfrac{1}{x^6}~=~\pu{cis}(6\theta)+\pu{cis}(-6\theta)~=~2 \pu{cos}\underbrace{\left(6\cdot \dfrac{\pi}{3}\right)}_{2\pi}~=~2$$$$\boxed{x^6 + \dfrac{1}{x^6}~=~2}$$