Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that  equation 
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots

I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
  $$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$
has two real roots but i don't have solution


HINT: you can write this equation:
$x^6+x^5-x^4-x^3=-x^2-x+1$
$y=x^6+x^5-x^4-x^3$
$y=-x^2-x+1$
and draw two functions noting that there are two points where two fnctions intersect.
