Here's an alternate construction of the invariant $\alpha$ that seems a little simpler (and, besides, gets used in proving the normal forms for $3$-forms in $6$-variables):

Let $S$ be a vector space over $\mathbb{C}$ of dimension $6$, and fix an isomorphism $\Lambda^6(S^*)= \mathbb{C}$ (i.e., choose a volume form). Then there is an induced natural isomorphism 
$$
S = S\otimes \Lambda^6(S^*) = \Lambda^5(S^*)  .\tag1
$$
Given $\phi\in\Lambda^3(S^*)$, define a mapping $J_\phi:S\to S$ by the rule
$$
J_\phi(s) = (\iota_s\phi)\wedge\phi
$$
where $\iota_s\phi\in\Lambda^2(S^*)$ is the interior product of $s\in S$ with $\phi$.  It is easy to see that the trace of $J_\phi\in \mathrm{End}(S,S) = S\otimes S^*$ vanishes identically.  However, if one sets
$$
\alpha(\phi) = \tfrac16\,\mathrm{tr}\bigl((J_\phi)^2\bigr),\tag2
$$
then one finds that $\alpha(\phi)$ does not vanish identically, and it is obviously a quartic polynomial in the coefficients of $\phi$.  In fact, one has the identity
$$
(J_\phi)^2 = \alpha(\phi)\,\mathrm{Id}_S\,,\tag3
$$
and this identity can be used to put $\phi$ in normal form with respect to the eigenspaces of $J_\phi$ when $\alpha(\phi)\not=0$.