Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.

$f(x)=x^8-x^5+x^2-x+1=0$ has no real root as $f(x)>0$ for all real values of $x$.
$$f(x)=x^3(x^3-1)+x(x-1)+1 >0, ~if~ x>1$$
$$f(x)=x^8+x^2(1-x^3)+(1-x) > ~if~x <1$$
And $f(1)=1>0$.
