  Given a nonzero integer $d$, we know by Hermite's Theorem that there exist
only finitely many cubic number fields of discriminant $d$. However, it can
happen that two non-isomorphic cubic fields have the same discriminant. It is
thus natural to ask whether there are natural refinements of the discriminant
which completely determine the isomorphism class of the cubic field. Here we
consider the trace form $q_K:\text{tr}_{K/\mathbb{Q}}(x^2)|_{O^{0}_{K}}$ as
such a refinement. For a cubic field of fundamental discriminant $d$ we show
the existence of an element $T_K$ in Bhargava's class group
$\Cl(\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}\otimes\mathbb{Z}^{2}; -3d)$ such that
$q_K$ is completely determined by $T_K$. By using one of Bhargava's composition
laws, we show that $q_K$ is a complete invariant whenever $K$ is totally real
and of fundamental discriminant
