prove $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b$ in an arbitrary field Let $F$ be an arbitrary field, I need to prove that $x^2 - x + 1$ divides $x^{10} - x^7 + x^4 + ax + b$ for some $a, b \in F$
The difficulty that I am currently facing is that since $F$ is an arbitrary field, $x^2 - x + 1$ might be irreducible over $F$ and so I cannot solve it by factoring $x^2 - x + 1$. I also try to use the division algorithm to prove it using contradiction by assuming $x^2 - x + 1$ don't divide it, and hopefully something would happen to the remainder but this leads to a dead end as well. So any help would be appreciated.

$$f(x)=x^4(x^6-x^3+1)+(ax+b)$$
Now reduce $f(x)$ by $x^2=x-1$ to have remainder $R(x)$ as
$$\implies R(x)=(x-1)^2[(x-1)^3-x(x-1)+1]+(ax+b)$$
$$\implies R(x)=(x-1-2x+1)[(x-1)(x^2-2x+1-x)+1])+ax+b$$
$$\implies R(x)=(x-1)^2[(x-1)(-2x)+1]+ax+b$$
$$\implies R(x)=(x-1)^2[-2(x-1)+2x+1]+ax+b$$
$$\implies R)x)=3(x-1)^2+ax+b \implies R(x)=-3x+ax+b$$
$$\implies R(x)=(a-3)x+b=0\implies a=3,b=0$$
