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


I'm  not sure of the general case; but for the first polynomial divide through by $x^3$, since $x\neq0$, and substitute $y=x-1/x$. You then obtain $y^3+y^2+2y+1=0$, which must have just one real root, in the interval $(-1,0).$  Call this root $y=a=x-1/x$. Then $x^2-ax-1=0$, which has a positive discriminant ($a^2+4$), implying two distinct real roots for $x$. I'm unsure if you can do something similar for the next polynomial...
