Hello!!!

I have applied a lot of times the euclidean division of $x^6-1$ with $x^2- \alpha^{a+1} (\alpha+1)x+ \alpha^{2a+3}, a \geq 0$, $\alpha$ a primitive $6$-th root of unity. But I don't get the right result...

We are over $\mathbb{F}_7$.

I got that $x^6-1=(x^2- \alpha^{a+1}(\alpha+1) x+ \alpha^{2a+3}) (x^4+ \alpha^{a+1} (\alpha+1) x^3+ \alpha^{2a+2}[(\alpha+1)^2- \alpha] x^2+ \alpha^{3a+3}(\alpha+1)[\alpha^2+1]x+ \alpha^{4a}[\alpha^4(\alpha+1)^2(\alpha^2+1)-[(\alpha+1)^2- \alpha]])+ \alpha^{5a}(\alpha+1)[\alpha^2+ \alpha+1+\alpha^5]x-2 \alpha^4-2\alpha^3-2 \alpha^8-\alpha^7+\alpha^4+\alpha^3-1$.

But the rest should be $0$. Have I done something wrong?