8. [25] Suppose $x$ is a fifth root of unity. Find, in radical form, all possible values of
$$
2 x+\frac{1}{1+x}+\frac{x}{1+x^{2}}+\frac{x^{2}}{1+x^{3}}+\frac{x^{3}}{1+x^{4}} .
$$
Solution: Note that
$$
\begin{array}{c}
\frac{x}{1+x^{2}}+\frac{x^{2}}{1+x^{3}}=\frac{x^{6}}{1+x^{2}}+\frac{x^{4}}{x^{2}+x^{5}}=\frac{x^{4}+x^{6}}{1+x^{2}}=x^{4}=\frac{1}{x}, \text { and } \\
\frac{1}{1+x}+\frac{x^{3}}{1+x^{4}}=\frac{x^{5}}{1+x}+\frac{x^{4}}{x+x^{5}}=\frac{x^{4}+x^{5}}{1+x}=x^{4}=\frac{1}{x} .
\end{array}
$$

Therefore, the sum is just $2 x+\frac{2}{x}$. If $x=1$, this is 4 . Otherwise, let $y=x+\frac{1}{x}$. Then $x$ satisfies
$$
0=1+x+x^{2}+x^{3}+x^{4}=\left(x^{2}+\frac{1}{x^{2}}+2\right)+\left(x+\frac{1}{x}\right)-1=y^{2}+y-1,
$$
so solving this quadratic yields $y=\frac{-1 \pm \sqrt{5}}{2}$, or $2 y=-1 \pm \sqrt{5}$. Since each value of $y$ can correspond to only 2 possible values of $x$, and there are 4 possible values of $x$ besides 1 , both of these values for $y$ are possible, which yields the answers, 4 and $-1 \pm \sqrt{5}$.