 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