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.

This equation has the same coefficients read backwards. 
There is a technique of solving such equations:
If the degree is odd, then $-1$ is a root, and dividing by $x+1$ gives you an even degree equation with the same property.
For even degree: divide by $x$ to the power half degree, and make the substitution $t=x+\frac{1}{x}$.
In this case we get
$$x^3+\frac{1}{x^3}-(x^2+\frac{1}{x^2})+x+\frac{1}{x}-1=0$$
we have
$$t=x+\frac{1}{x} \\
t^2=x^2+\frac{1}{x^2}+2\\
t^3=x^3+\frac{1}{x^3}+3t$$
Therefore, your equation becomes
$$t^3-3t-t^2+2+t-1=0\\
t^3-t^2-2t+1=0$$
You can solve this by using the cubic formula, and then solve the corresponding quadratics.
