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

 A: 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.
A: 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...
A: since $ f(\pm \infty) = \infty, f(0) = -1$ shows that $f$ has one positive root and one negative root. so $f$ has at least two real roots.
A: 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.
A: 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.
