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.

You got the right expression. $x^7+1=0$
The roots of the equation will be like $$x= \cos (\frac{2k\pi}{7})+ i\sin (\frac{2k\pi}{7}) , 0\le k\le6$$
Note: Stress on be like, this formula will not give you the exact roots. You have to change it a bit to account for the roots of negative unity.
EDIT:  Complete solution follows:
$$x^7+1=0$$ or,
$$(x+1)(x^6-x^5+x^4-x^3+x^2-x+1)=0$$
This implies your equation has exactly 1 real root and 3 pairs of complex conjugate roots. 
Hence now I can write 
$$x^7+1=0$$ or,
$$x^7=-1$$ or,
$$x^7=\cos \pi + i\sin \pi = \cos (2k+1)\pi + i\sin (2k+1)\pi , 0 \le k\le 6 $$ or,
$$x=[\cos (2k+1)\pi + i\sin (2k+1)\pi]^{\frac{1}{7}} , 0 \le k\le 6$$ or,
$$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6$$
From the comment by Macavity: However,you have a spurious root included - $k=3$ . By multiplying by $x+1$ you introduced this root which is not a root of the original polynomial. So there are no real roots for the polynomial, only complex ones. Hence the final solutions are as follows: $$x=\cos \frac{(2k+1)\pi}{7}+ i\sin \frac{(2k+1)\pi}{7} , 0 \le k\le 6 \,\ \text{and} \,\ k \not = 3$$
