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 $\alpha^3 = \alpha^2+1$?