Decompose polynomial of degree 3 in a sum of cubed orthogonal linear forms

Question

Given is a homogenous polynomial $p$ of degree $3$ with $n$ terms where the coefficients can only be $\pm1$. The polynomial has $m$ variables and in each term the variables have degree $1$, i.e.,

$$p(x_1,x_2,\dots,x_m)=\sum\limits_{l=1}^n s_l\; x_{i}\; x_{j}\; x_{k},$$

where $s_l = \pm 1$, and $i, j, k \in \{1, \ldots, m\}$ and $i \ne j \ne k$ for a given $l$. A proper example would be the polynomial $$x_1 x_2 x_5-x_2x_3x_4-x_1x_3x_5$$ with $n=3$, $m=5$, $s_1=1$, $s_2=-1$, $s_3=-1$.

Let's assume there is a decomposition of $p$ by a sum of cubes of orthogonal linear forms with coefficient vectors ${\bf a}_w=(a_{w1},a_{w2},\ldots,a_{wm})$, i.e.

$$p(x_1,x_2,\dots,x_m)=\sum\limits_{w=1}^r \langle{\bf a}_w{\bf, x}\rangle^3,$$

where

$$\langle{\bf a}_u, {\bf a}_v\rangle\begin{cases}=0\text{ if } u\ne v\\\ne0\text{ if }u= v,\end{cases}$$

and due to requested orthogonality $r\le m$. Here, ${\bf x}=(x_1,x_2,\dots,x_m)$ indicates the vector of the variables of the polynomial, and $\langle\cdot,\cdot\rangle$ is the dot product.

In case such a special polynomial is decomposable, I would know if there is a method to find the vectors ${\bf a}_w$? I am only aware how to solve this if the degree of the polynomial would be $2$ but not if the degree is $3$.

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Simple non-orthogonal example

For the polynomial $x_1x_2x_3$ we have $n=1$, $m=3$, $s_1=1$ and concerning this post and this paper$\color{red}{^\star}$ it can be decomposed in a sum of four cubes of linear forms, however the coefficient vectors are not mutually orthogonal.

$$x_1 x_2 x_3=\left(\frac{x_1}{2 \sqrt[3]{3}}+\frac{x_2}{2 \sqrt[3]{3}}+\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(\frac{x_1}{2 \sqrt[3]{3}}-\frac{x_2}{2 \sqrt[3]{3}}-\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(-\frac{x_1}{2 \sqrt[3]{3}}+\frac{x_2}{2 \sqrt[3]{3}}-\frac{x_3}{2 \sqrt[3]{3}}\right)^3+\left(-\frac{x_1}{2 \sqrt[3]{3}}-\frac{x_2}{2 \sqrt[3]{3}}+\frac{x_3}{2 \sqrt[3]{3}}\right)^3,$$

with coefficient vectors

$${\bf a}_1=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}1\\1\\1\end{pmatrix}, \quad {\bf a}_2=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}1\\-1\\-1\end{pmatrix}, \quad {\bf a}_3=\frac{1}{2\sqrt[3]{3}}\begin{pmatrix}-1\\1\\-1\end{pmatrix}, \quad {\bf a}_4=\frac{1}{2\sqrt[3] {3}}\begin{pmatrix}-1\\-1\\1\end{pmatrix}$$

I am not aware if for this polynomial exists a decomposition of orthogonal coefficients. Also I am not aware of another example where the coefficients are orthogonal.

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Reference

$\color{red}{\star}$ Boris Reichstein, On expressing a cubic form as a sum of cubes of linear forms, Linear Algebra and its Applications, Volume 86, February 1987.

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Related question: Link

Answer 1

For your example $x_1 x_2 x_3$, it seems there is no decomposition into cubes of three linear forms. I used Maple to look at the system of $10$ equations in $9$ variables $a_{ij}$ obtained by taking the coefficients of $x_1 x_2 x_3 - \sum_{i=1}^3 (a_{i1} x_1 + a_{i2} x_2 + a_{i3} x_3)^3 $; a Groebner basis was $\{1\}$, indicating that the system is inconsistent.

EDIT: Maple code is:

eqs:= convert(collect(x[1]*x[2]*x[3]-add((a[i,1]*x[1]+a[i,2]*x[2]+a[i,3]*x[3])^3,i=1..3),
[x[1],x[2],x[3]],distributed),set);
G:= subs(seq(x[i]=1,i=1..3),eqs);
Groebner:-Basis(G,tdeg(seq(seq(a[i,j],i=1..3),j=1..3)));