So, given this function: $$L(A) = \sum_{IJKLM} \alpha^J_L \beta^K_M \gamma^{ILM} A_I A_J A_K$$ We desire to find the matrices $\alpha$ and $\beta$, where all elements of $\alpha$, $\beta$ and $\gamma$ are $\in \{0, 1, w, w^2\}$ and $\gamma$ is already fully known. Where $w$ is the principle root of unity and hence has the properties that $w^3 = 1$ and $1 + w + w^2 = 0$. I,J,K,L are all over the same set (with 10 elements), and M is a set with 6 elements (19 if I made a false assumption).
The 10 element set is: {(0,0), (1,1), (2,2), (3,3), (0,1), (0,2), (0,3), (2,3), (1,3), (1,2)}
The 6 element set is: {(0,0,1,1), (0,0,2,2), (0,0,3,3), (1,1,2,2), (1,1,3,3), (2,2,3,3)}
$\gamma$ has the property that if an index is repeated more than 2 times it is zero. Otherwise it is $w$ raised to power $(c\mod 3)$, where $c$ is the number of swaps required to sort the indices in ascending order. Essentially a brother to the standard Levi-Civita symbol.
Next, we define this polynomial: $$J_3(a,b,c) = (a + b + c)(a + wb + w^2c)(a + w^2b + wc) = a^3 + b^3 + c^3 - 3abc$$
If we set: $$F = [V_0, V_1, V_2, V_3, X_1, X_2, X_3, Y_1, Y_2, Y_3]$$
Then the current working assumption is that we require: $$L(F) = J_3(V_0, V_1, X_1) + J_3(V_0, V_2, X_2) + J_3(V_0, V_3, X_3) + J_3(V_2, V_3, Y_1) + J_3(V_1, V_3, Y_2) + J_3(V_1, V_2, Y_3)$$
There may be more non-zero terms in $L(F)$, I'm not 100% certain, but it is unlikely. If there are more terms, they must also use $J_3()$. Now, the challenge is to find the $\alpha$ and $\beta$ matrices. I think I have correctly narrowed it down from 160 to a total of 60 free parameters $\in \{1, w, w^2\}$ by finding the zero terms. And I have working code in SageMath/Jupyter/sympy that implements $L(A)$, and normalizes the roots of unity terms, including applying $1 + w + w^2 = 0$.
Using sympy, and isolating the coefficients of terms such as $V_0^3$ or $V_1^3$ or $V_0 V_1 X_1$ or $V_0 X_1^2$ or $V_0^2 V_1$ then gives me a large collection of equations of form: $$a_{01} b_{05} + a_{02} b_{04} w + a_{03} b_{03} w^{2} = 3$$ $$a_{10} b_{15} w + a_{12} b_{12} w^{2} + a_{13} b_{11} = 3$$ $$a_{04} b_{15} w + a_{14} b_{05} w + a_{40} b_{05} w + a_{41} b_{15} + w^{2} + w = - 3$$ $$a_{04} + a_{41} w^{2} + b_{05} w = 0$$ $$a_{01} b_{15} + a_{02} b_{02} w^{2} + a_{02} b_{14} w + a_{03} b_{01} + a_{03} b_{13} w^{2} + a_{12} b_{04} w + a_{13} b_{03} w^{2} + b_{05} w^{2} + b_{05} = 0$$
At which point I'm stumped. At least our equations have the right number of terms though, so a solution is at least plausible. Eg, to get $-3$, we need $6$ terms (which makes use of $w + w^2 = -1$). To get $0$ we need a multiple of $3$ terms (which makes use of $1 + w + w^2 = 0$).
For reference, here are my current $\alpha$ and $\beta$ matrices:
alpha[0] = [w^2, a01, a02, a03, a04, a05, a06, 0, 0, 0]
alpha[1] = [a10, w^2, a12, a13, a14, 0, 0, 0, a18, a19]
alpha[2] = [a20, a21, w^2, a23, 0, a25, 0, a27, 0, a29]
alpha[3] = [a30, a31, a32, w^2, 0, 0, a36, a37, a38, 0]
alpha[4] = [a40, a41, 0, 0, 1, 0, 0, 0, 0, 0]
alpha[5] = [a50, 0, a52, 0, 0, 1, 0, 0, 0, 0]
alpha[6] = [a60, 0, 0, a63, 0, 0, 1, 0, 0, 0]
alpha[7] = [0, 0, a72, a73, 0, 0, 0, 1, 0, 0]
alpha[8] = [0, a81, 0, a83, 0, 0, 0, 0, 1, 0]
alpha[9] = [0, a91, a92, 0, 0, 0, 0, 0, 0, 1]
beta[0] = [b00, b01, b02, b03, b04, b05]
beta[1] = [b10, b11, b12, b13, b14, b15]
beta[2] = [b20, b21, b22, b23, b24, b25]
beta[3] = [b30, b31, b32, b33, b34, b35]
beta[4] = [0, 0, 0, 0, 0, w^2]
beta[5] = [0, 0, 0, 0, w, 0]
beta[6] = [0, 0, 0, 1, 0, 0]
beta[7] = [w, 0, 0, 0, 0, 0]
beta[8] = [0, w^2, 0, 0, 0, 0]
beta[9] = [0, 0, 1, 0, 0, 0]
Update, so using $V^3_\mu = 3$ it is possible to eliminate 12 more parameters:
alpha[0] = [w^2, 1/b05, w^2/b04, w/b03, a04, a05, a06, 0, 0, 0]
alpha[1] = [w^2/b15, w^2, w/b12, 1/b11, a14, 0, 0, 0, a18, a19]
alpha[2] = [w/b24, 1/b22, w^2, w^2/b20, 0, a25, 0, a27, 0, a29]
alpha[3] = [1/b33, w^2/b31, w/b30, w^2, 0, 0, a36, a37, a38, 0]
alpha[4] = [a40, a41, 0, 0, 1, 0, 0, 0, 0, 0]
alpha[5] = [a50, 0, a52, 0, 0, 1, 0, 0, 0, 0]
alpha[6] = [a60, 0, 0, a63, 0, 0, 1, 0, 0, 0]
alpha[7] = [0, 0, a72, a73, 0, 0, 0, 1, 0, 0]
alpha[8] = [0, a81, 0, a83, 0, 0, 0, 0, 1, 0]
alpha[9] = [0, a91, a92, 0, 0, 0, 0, 0, 0, 1]