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


Let $f(x)=x^6+x^5-x^4-x^3+x^2+x-1=0$. Then $f(0)=-1$. Note for $x>0$
\begin{eqnarray}
f'(x)&=&(6x^5+2x)+(5x^4+1)-4x^3-3x^2\\
&\ge& 4\sqrt3x^3+2\sqrt5x^2-4x^3-3x^2\\
&=&(4\sqrt3-4)x^3+(2\sqrt5-3)x^2\\
&>&0
\end{eqnarray}
hance $f(x)$ is strictly increasing. Noting that $\lim_{x\to\infty}f(x)=\infty$, we have a unique root in $(0,\infty)$. Similarly, there is a unique root in $(-\infty,0)$. Thus $f(x)$ has only two real roots in $(-\infty,\infty)$. Using the same argument for other function, you can get the answer.
