<s>[INST] 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 $x^8-x^7+x^5-x^4+x^3-x+1$ ?
 [/INST] 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.]
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