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 $x^8-x^7+x^5-x^4+x^3-x+1$ ?
 A: 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 $\{e^{\frac{2ni\pi}{5}}\mid n\in\mathbb Z \}$.
A: The direct solution has been posted already, so the following is just to provide an elementary (but quite laborious) alternative that involves nothing more than solving quadratics, as an answer to this part of OP's question.

how am I suppose to finding roots in $\mathbb{C}$ for $x^8-x^7+x^5-x^4+x^3-x+1$

This is a palindromic polynomial, and the degree can be halved with the substitution $\,x+\frac{1}{x}=u\,$. After routine calculations, this results in $\,u^4 - u^3 - 4 u^2 + 4 u + 1 = 0\,$.
The substitution $\,u-1=v\,$ then gives $\,v^4 + 3 v^3 - v^2 - 3 v + 1=0\,$, which is again a palindromic polynomial, and finally $\,v - \frac{1}{v} = w\,$ gives the quadratic $\,w^2 + 3 w + 1=0\,$.
Each of the two roots $\,w\,$ of the latter gives two $\,v\,$ roots found by solving $\,v^2 - w v - 1 = 0\,$, then each of the four $\,v\,$ roots gives one root $\,u = v+1\,$, and each of the four $\,u\,$ roots gives two $\,x\,$ roots found by solving $\,x^2-ux+1=0\,$.
A: We might also pretend we don't know about cyclotomic polynomials and try constructing the roots of the equation from properties of complex numbers.  From the equation $ \ z^{10} + z^5 + 1 \ = \ 0 $ $ \ \Rightarrow \ z^{10} + z^5 \ = \ -1 \ \ , $ we observe that $ \ w \ = \ z^5 \ $ and $ \ w^2 \ $ sum to a real number, so we have $ \ w^2 \ = \ \overline{w} \ \ . $  If we write $ \  w \ = \ \alpha + \beta·i \ \ , $ then this equation tells us that
$  (\alpha^2 \ - \ \beta^2) \ + \ 2·\alpha·\beta·i \ \ = \ \ \alpha \ - \ \beta·i $
$\Rightarrow \ \ \alpha \ = \ -\frac12 \ , \ \beta \ \neq \ 0 \ \ $ by equating imaginary parts (we know $ \ w \ $ is complex since $ \ w^2 + w + 1 \ = \ 0 \ $ has a negative discriminant)
$ \Rightarrow \ \  \left(-\frac12 \right)^2 \ - \ \beta^2 \ = \ -\frac12 \ \Rightarrow \ \beta^2 \ = \ \frac34 \ \ . $
We do this to satisfy ourselves that $ \ |w| \ = \ 1 \ \ . $  Thus, $ \ w \ = \ -\frac12 + \frac{\sqrt3}{2}·i \ = \ e^{ \ i·2 \pi/3} \ \ ,  $ $ w^2 \ = \ \overline{w} \ = \ -\frac12 - \frac{\sqrt3}{2}·i \ = \ e^{ \ i·4 \pi/3} \ \ . $
The ten roots of the original equation are then the five complex fifth-roots each of $ \ w \ $ and $ \ \overline{w} \ \ , $
$$ z \ \ = \ \ \large{e^{ \ i·\left(\frac{2 \pi}{15} + \frac{2·k·\pi}{3} \right)} } \ \ = \ \ e^{ \ i·\left(\frac{(2 + 6k) \pi}{15} \right)} \ \ , \ \ \large{e^{ \ i·\left(\frac{4 \pi}{15} + \frac{2·k·\pi}{3} \right)} } \ \ = \ \ e^{ \ i·\left(\frac{(4 + 6k) \pi}{15} \right)} \ \ , $$
or $$ z \ \ = \ \ \large{e^{ \ i·\frac{2 \pi}{15} }  \ , \ e^{ \ i·\frac{4 \pi}{15} }   \ , \ e^{ \ i·\frac{8 \pi}{15} }  \ , \ e^{ \ i·\frac{10 \pi}{15} } \ = \  e^{ \ i·\frac{2 \pi}{3} }  \ , \ e^{ \ i·\frac{14 \pi}{15} }  \ , \ e^{ \ i·\frac{16 \pi}{15} }   \ ,} $$ $$ \large{e^{ \ i·\frac{20 \pi}{15} }  \ = \  e^{ \ i·\frac{4 \pi}{3} }  \ , \ e^{ \ i·\frac{22 \pi}{15} }   \ , \ e^{ \ i·\frac{26 \pi}{15} }  \ , \ e^{ \ i·\frac{28 \pi}{15} }  \ \ . } $$
On the Argand diagram, these roots marks the vertices of two regular pentagons rotated 24º relative to one another.  Note that the fifth-roots of $ \ 1 \ $ are excluded from the set of implied fifteenth-roots of $ \ 1 \ \ . $
[I acknowledge the much nicer and more sophisticated arguments by MH.Lee and dxiv ; I was interested here in what the least means employed to answer the question might be.]
