Solution
= Solution
The cubic $x^3+x+1$ has no zero in $\mathbb F_2$, so is irreducible. Its quotient field has eight elements. Every nonzero element of that field satisfies $x^7=1$ by the order of its multiplicative <group>, so the polynomial splits there. It does not already split in the base field; a root of that irreducible cubic, which divides $x^7-1$, is needed. Hence
$$
\boxed{K\cong\mathbb F_8,\qquad[K:\mathbb F_2]=3}.
$$