prove that the polynomial $x^8 -x^5 +x^2 -x +1$ is positive for all real values of x is there any factorization possible of the above expression or can it be shown that is a a sum of two or three squares?
i tried various factorizations but none of them were conclusive.

$$x^8-x^5+x^2-x+1 = \frac{1}{2}\bigg[2x^8-2x^5+2x^2-2x+2\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^8-2x^5+x^2)+(x^2-2x+1)+1\bigg]$$
So $$ = \frac{1}{2}\bigg[x^8+(x^4-x)^2+(x-1)^2+1\bigg]>0\forall x \in \mathbb{R}.$$
Added:: For $x=0,x^8-x^5+x^2-x+1>0$
Using Arithmetic Geometric Inequality $(x\neq 0)$
$$\frac{x^8}{2}+\frac{x^2}{2}\geq |x^5|\geq x^5$$
$$\frac{x^2}{2}+\frac{1}{2}\geq |x|\geq x$$
So $$\frac{x^8}{2}+\frac{x^2}{2}-x^5+\frac{x^2}{2}-x+\frac{1}{2}+\frac{1}{2}>0\forall x \in \mathbb{R}$$
