, 2014

Solution:
Since $x + x^{-1}$ is a root,
$$\begin{aligned} 0 & = \left(x + x^{-1}\right)^{3} + \left(x + x^{-1}\right)^{2} - 2\left(x + x^{-1}\right) - 1 \\ & = x^{3} + x^{-3} + 3x + 3x^{-1} + x^{2} + 2 + x^{-2} - 2x - 2x^{-1} - 1 \\ & = x^{3} + x^{-3} + x^{2} + x^{-2} + x + x^{-1} + 1 \\ & = x^{-3}\left(1 + x + x^{2} + \cdots + x^{6}\right). \end{aligned}$$
Since $x \neq 0$, the above equality holds only if $x$ is a primitive seventh root of unity, i.e. $x^{7} = 1$ and $x \neq 1$. Therefore, the only possible value of $x^{7} + x^{-7}$ is $1 + 1 = 2$.