= Solution
Over $\mathbb F_2$, subtraction equals addition. The polynomial $X^7-1$ has the factorization
$$
\boxed{
X^7-1=(X+1)(X^3+X+1)(X^3+X^2+1).}
$$
The two cubic factors have no root in $\mathbb F_2$, so they are irreducible, and they are reciprocal to one another. Put
$$
f=X^3+X+1,
\qquad
f^*=X^3+X^2+1.
$$
Every cyclic code corresponds to one monic divisor of $X^7-1$. The complete list is
$$
\begin{array}{c|c|c}
g(X)&\text{parameters}&\text{code}\\ \hline
1 &[7,7,1]&\text{whole-space linear code}\\
X+1 &[7,6,2]&\text{even-weight binary code}\\
f &[7,4,3]&\text{Hamming code}\\
f^* &[7,4,3]&\text{reversed Hamming code}\\
(X+1)f &[7,3,4]&\text{binary simplex code}\\
(X+1)f^* &[7,3,4]&\text{reversed binary simplex code}\\
ff^*=1+X+\cdots+X^6 &[7,1,7]&\text{binary repetition code}\\
(X+1)ff^*=X^7-1 &[7,0]&\text{zero code}.
\end{array}
$$
Thus the two cubic generators give the two cyclic coordinate orientations of Hamming's $[7,4,3]$ code. Their duals have generators $(X+1)f$ and $(X+1)f^*$ and are the $[7,3,4]$ simplex codes. The remaining four are the familiar whole-space, even-weight, repetition, and zero codes. This is the <binary cyclic codes of length seven>[complete classification of binary cyclic codes of length seven].
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