Let $\alpha$ be a root of $x^2 + x + 2 = 0$ in $F_3[x]$. I am asked to show that $x^3 + x + 1$ has roots $\alpha$, $\alpha^2$ and $\alpha^4$.
I started by observing that $\alpha^2 + \alpha + 2 = 0 \implies \alpha^2 = -\alpha - 2 = 2\alpha + 1$. Then $\alpha^3 + \alpha + 1$ is just $\alpha(2\alpha + 1) + \alpha + 1 = 2(2\alpha+1) + \alpha + \alpha + 1 = 6\alpha+3 = 0$.
But when I go to do something similar for $\alpha^2$, I run into some trouble. I know that $\alpha^6 = \alpha+2$ from a previous problem (which I got correct, and it is easily verified). But when I make this substitution into $\alpha^6 + \alpha^2 + 1$, I get $(\alpha + 2) + (2\alpha + 1) + 1 = 1$, not zero. Where am I going wrong?