= Solution
Over $\mathbb F_2$, the cubic takes value one at both zero and one, so it has no root. A reducible cubic over a field has a linear factor and therefore a root; hence this cubic is irreducible. Adjoin a root:
$$
K=\mathbb F_2[\alpha],\qquad\alpha^3+\alpha+1=0.
$$
This extension has <basis> $1,\alpha,\alpha^2$, degree three and eight elements. Squaring preserves <polynomial> equations with coefficients in $\mathbb F_2$, so $\alpha^2$ and $\alpha^4$ are roots too. The nonzero element $\alpha$ is not one; the multiplicative group has prime order seven, so $\alpha$ has order seven. The exponents $1,2,4$ are distinct modulo seven, making these three roots distinct. Thus they are all the roots and already lie in $K$. Therefore
$$
\boxed{\text{The splitting field is }\mathbb F_8=\mathbb F_2[X]/(X^3+X+1),\quad[K:\mathbb F_2]=3.}
$$
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