Can anyone point me in the right direction with the following problem?
Given that $$GF(8)=\frac{Z_{2}}{x^3+x^2+1}= \frac{Z_{2}}{x^3+x+1}$$
Find $\beta$ as a function of $\alpha$ , where $\alpha$ is a root of $x^3+x+1$ and $\beta$ is a root of $x^3+x^2+1$
On
Let $\alpha$ be a root of $x^3+x+1$. Then a general element of $GF(8)$ has the form $a+b\alpha + c\alpha^2$ for some $a,b,c \in GF(2)$. Need to solve for for $a,b,c$ in $(a+b\alpha + c\alpha^2)^3+(a+b\alpha + c\alpha^2)^2+1 = 0$ Recall that $u^2=u$ for any $u \in GF(2)$ and $2u=0$ for any $u \in GF(8)$. After simplifying you'll find $$ (1+b+c+bc) + (b+c+ab+ac)\alpha + (b+ab)\alpha^2 = 0$$ from which you can determine that $\beta$ can be taken to be $1+\alpha, 1+\alpha^2, $ or $1+\alpha+\alpha^2$
Brief answer:
As one polynomial is just the reverse of the other, the roots of one are the inverses of the roots of the other. Hence we may take $\beta=\alpha^{-1}$
Here is the complete story:
One can find the three possibilities this way.
Neither $0$ nor $1$ is a root of either $X^3+X^2+1$ or $X^3+X+1$; and it is trivial to see that these polynomials are coprime and so have no common root.
The order of the multiplicative group of $\mathbb{F}_8$ is seven and so is cyclic of that order.
If $\alpha$ is a root of $X^3+X^2+1$ so is $\alpha^2$ since $\alpha^6+\alpha^4+1=(\alpha^3+\alpha^2+1)^2=0$. Hence the roots of this polynomial are $\alpha, \alpha^2, \alpha^4(=(\alpha^{2})^2)$ and these are distinct as the multiplicative order of $\alpha$ is $7$.
In the same way the roots of $X^3+X+1$ are $\beta,\beta^2,\beta^4$. That accounts for all the elements of $\mathbb{F}_8$.
Therefore $\beta,\beta^2,\beta^4$ must be the "missing" powers of $\alpha$, namely $\alpha^3,\alpha^5,\alpha^6$: or more elegantly $\alpha^{-1}, \alpha^{-2}, \alpha^{-4}$.
So $\beta$ is one of $\alpha^{-1}, \alpha^{-2}, \alpha^{-4}$.