<s>[INST] Express each power of the root $\alpha$ of $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle}$ as linear combinations of $1, \alpha$ and $\alpha^2$ There are $8$ elements in $\frac{\mathbb Z_2[x]}{\langle x^3+x^2+1\rangle} = GF(8)$ 
and this generates the set $\{0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1\}$
We're required to express $\alpha^1$ all the way up to $\alpha^7$ as linear combinations of $1, \alpha$ and $\alpha^2$
$\alpha^1 = \alpha$
$\alpha^2 = \alpha^2$
$\alpha^3 = \alpha^2 + 1$
$\alpha^4 = \alpha^2+\alpha+1$
$\alpha^5 = \alpha+1$
$\alpha^6 = \alpha^2+\alpha$
$\alpha^7 = 1$
I'm really not seeing where these combinations are coming from. Why is $\alpha^3 = \alpha^2+1$?
 [/INST] If $\alpha$ is a root of $x^3+x^2+1$, then $\alpha^3+\alpha^2+1=0$, so $\alpha^3=\alpha^2+1$ (since $1=-1$ over $\mathbb{Z}_2$).
From here: 
$$\begin{array}{l} \alpha^4=\alpha\alpha^3=\alpha(\alpha^2+1)=\alpha^3+\alpha=\alpha^2+\alpha+1\\
\alpha^5=\alpha\alpha^4=\alpha^3+\alpha^2+\alpha=\alpha^2+1+\alpha^2+\alpha=\alpha+1\\
\alpha^6=\alpha\alpha^5=\alpha^2+\alpha\\
\alpha^7=\alpha\alpha^6=\alpha^3+\alpha^2=\alpha^2+1+\alpha^2=1 \end{array}$$
(using $1+1=0$ over $\mathbb{Z}_2$)
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