Give a complete non-redundant list of elements $x\in\mathbb{C}$ such that $x^{10}+x^5+1=0$ Since, $x^{10}+x^5+1 = (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)=0$.
I know that $x^2+x+1=0$ has root $x=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
How am I supposed to find the roots in $\mathbb{C}$ for $x^8-x^7+x^5-x^4+x^3-x+1$ ?

Since $x^{10}+x^5+1=\frac{x^{15}-1}{x^5-1}$, root of $x^{10}+x^5+1=0$ must be root of $x^{15}-1=0$, but not root of $x^5-1=0$. $x^5-1=0$ and $x^{10}+x^5+1=0$ has no common root.
So, the zero set of $x^{10}+x^5+1$ is...

 $$\{e^{\frac{2ni\pi}{15}}\mid 3\not\mid n\in\mathbb Z \}.$$

Because the zero set of $x^{15}-1=0$ is $\{e^{\frac{2ni\pi}{15}}\mid n\in\mathbb Z \}$, and the zero set of $x^5-1=0$ is $\{e^{\frac{2ni\pi}{5}}\mid n\in\mathbb Z \}$.
