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 $x^8-x^7+x^5-x^4+x^3-x+1$ ?

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 $\,x^2-ux+1=0\,$.
