Question: 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.

Answer: 
$$2(x^8-x^5+x^2-x+1)=x^8+(x^4-x)^2+(x-1)^2+1.$$
For $x<1$ we have $$x^8+\underbrace{(1-x)}\_{>0}+x^2\underbrace{(1-x^3)}\_{>0}>0$$

and for $x>1$ we have $$x^5\underbrace{(x^3-1)}\_{>0}+x\underbrace{(x-1)}\_{>0}+1>0$$