How to solve $x^6-x^5+x^4-x^3+x^2-x+1=0$? Can anyone tell me how to solve this?
$x^6-x^5+x^4-x^3+x^2-x+1=0$
What I got to was $x^7+1=0$.  
Thanks in advance.

$$x^6-x^5+x^4-x^3+x^2-x+1=0 \Longleftrightarrow$$
$$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0 \Longleftrightarrow$$
$$x^7+1=0 \Longleftrightarrow$$

This introduces the extraneous root of $x=-1$, so from now on we assume that $x\ne -1$:

$$x^7+1=0 \Longleftrightarrow$$
$$x^7=-1 \Longleftrightarrow$$
$$x^7=e^{\pi i} \Longleftrightarrow$$
$$x=\left(e^{(\pi+2\pi k) i}\right)^{\frac{1}{7}} \Longleftrightarrow$$
$$x=e^{\frac{1}{7}(\pi+2\pi k) i}$$
With $k\in\mathbb{Z}$ and $k:0-6$
