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


The first equation is dealt with well by @user140591's answer. The second equation, $x^6-x^5+x^4+x^3-x^2-x+1=0,$ has no real roots. To prove this, we first show that it has no negative roots. This can be seen by writing the equation as $$x=\frac{x^6+x^4-x^2+1}{x^4-x^2+1}\qquad\qquad\qquad\quad$$
$$=\dfrac43\frac{3(x^3-x)^2+(3x^2-1)^2+2}{(2x^2-1)^2+3}.$$In the first displayed equation, write $x^2=t$ and divide to get $$\sqrt t=t+2-\frac{4}{(2t-1)^2+3}.$$To show that the RHS exceeds the LHS for $t\geqslant0$, it is enough to show that it does so when the negative term is replaced by its least value, namely $-\frac43$, and the LHS is replaced by $\frac12t+\frac12$, which upper-bounds it. This is easily seen to be the case.
