Prove the splitting field of $\displaystyle x^3+x^2+1\in Z_{2}[x]$ is a radical extension...
when it comes to $\displaystyle Z_{2}$,I am a little bit confused. Since many theorem is based on $\displaystyle charF=0$..
The theorem about determining if something is a radical extension (that is, the polynomial is solvable) is about charachteristic zero fields. However, the definition of radical extensions is more general then that.
Remember, $\displaystyle K/F$ is a radical extension (by definition) iff there exists $\displaystyle a_1,...,a_n\in K$ and $\displaystyle e_1,...,e_n\geq 1$ such that $\displaystyle F(a_1,...,a_n) = K$, $\displaystyle a_j^{e_j} \in F(a_1,...,a_{j-1})$.
So construct the splitting field of this and argue that you can write it in the above form.
Notice that $\displaystyle x^3+x^2+1$ is irreducible over $\displaystyle F = \mathbb{Z}_2[x]$. Therefore, as you know, there exists an extension field $\displaystyle K$ which has $\displaystyle \alpha\in K$ that solves this polynomial. Therefore, $\displaystyle \alpha^3 = \alpha^2 + 1$. Now $\displaystyle x^3 + x^2 + 1 = (x+\alpha)(x^2 + (\alpha+1)x+\alpha(\alpha+1))$. You need to ask now whether $\displaystyle x^2+(\alpha+1)x+\alpha(\alpha+1)$ has a zero in $\displaystyle F(\alpha)$. Sadly, it does not, this can be confirmed by checking $\displaystyle a\alpha^2+b\alpha + c$ where $\displaystyle a,b,c\in \{0,1\}$. Thus, $\displaystyle F(\alpha)$ is not the splitting field over $\displaystyle x^3+x^2+1$. However, we know there exists $\displaystyle L/F(\alpha)$ such that there is $\displaystyle \beta \in L$ which solves $\displaystyle x^2+(\alpha+1)x+\alpha(\alpha +1 ) \in F(\alpha)[x]$. The extension field $\displaystyle F(\alpha,\beta)$ will therefore become the splitting field over $\displaystyle F$. Now it remains to argue this satisfies the conditions of being a radical extension.
this is maybe an easier way: let $\displaystyle p(x)=x^3+x^2+1 \in \mathbb{F}_2[x].$ see that if $\displaystyle p(\alpha)=0,$ then $\displaystyle \alpha^7 = 1 \in \mathbb{F}_2.$ thus the splitting field of $\displaystyle p(x)$ is a radical extension of $\displaystyle \mathbb{F}_2.$
But $\displaystyle F(\alpha)$ is not the splitting field.
You need to show that $\displaystyle F(\alpha,\beta)$ satisfies $\displaystyle \beta^{n_1} \in F(\alpha)$ and $\displaystyle \alpha^{n_2} \in F$.
You have shown that $\displaystyle n_2=7$ works so it still remains to find $\displaystyle n_1$.