Find the splitting field $E$ of $f(x)=x^3+1$ over $F= \mathbb{Z_2}$ and compute $[E:F]$.
First we find the roots of the polynomial $f(x)=x^3+1$
$f(x)=(x+1)(x^2-x+1)=0$ gives $x=-1, \frac{1\pm \sqrt{3}i}{2} $
Hence, $E=F(-1, \frac{1\pm \sqrt{3}i}{2})=F(\frac{1 + \sqrt{3}i}{2})$
since $[-1]=[1] \in F $ and $\frac{1 - \sqrt{3}i}{2}=1+\frac{1 - \sqrt{3}i}{2}$
Note that $x^2-x+1$ is irreducible in $F$ since there are only two linear factors $x, x+1$ in $F$.
Then $[E:F]=[F(\frac{1 + \sqrt{3}i}{2}): F]$=2
Do you think my argument is correct?
You did well until the first descomposition: $$f(x)=(x+1)(x^2-x+1)$$ or, since $1=-1$ in this field: $$f(x)=(x+1)(x^2+x+1)$$
Now, since the characteristic of $F$ is $2$ you can not apply the formula for solving second degree equations.
Observe instead that $0$ and $1$ are not roots of $x^2+x+1$, so every root of this polynomial is outside $F$. Since they are roots of a second degree equation, this polynomial is the minimum polynomial over $F$ of these roots. The belong to a normal extension of degree $2$, that is, $E=\Bbb F_4$.
This field is the only field (up to iso) with $4$ elements.
$$\Bbb F_4=\{0,1,\alpha,\alpha+1\}$$
where $\alpha^2+\alpha+1=0$.
Note that also $(\alpha+1)^2+(\alpha+1)+1=\alpha^2+2\alpha+1+\alpha+1+1=\alpha^2+\alpha+1=0$