= Solution
The cubic has no <root> in $\mathbb F_2$, since its values at $0$ and $1$ are both $1$. A cubic over a <field> is <irreducible polynomial>[irreducible] precisely when it has no <root>. Consequently
$$
K=\mathbb F_2[\alpha],\qquad \alpha^3+\alpha+1=0,
$$
is a <field extension> of <degree of a field extension>[degree] $3$, with eight elements. The <Frobenius endomorphism> $x\mapsto x^2$ takes a <root> of a <polynomial> over $\mathbb F_2$ to another <root>, so the roots are $\alpha,\alpha^2,\alpha^4$. They are distinct: equality between any two would make the <Frobenius endomorphism> orbit have length less than $3$, contradicting the <minimal polynomial> of $\alpha$. They all lie in $K$, and one of them generates $K$. Hence
$$
\boxed{K\cong\mathbb F_8,\qquad [K:\mathbb F_2]=3.}
$$
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