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This is a fairly straightforward question, and I am hoping a definitive answer exists.

Does there exist a quadratic form $A \in \mathbb{C}[x_1, x_2, x_3, x_4]$ and a cubic form $B \in \mathbb{C}[x_1, x_2, x_3, x_4]$ such that

$$D = 4A^3 + 27B^2 = L_1 \cdots L_6,$$

where the $L_i$'s are distinct (pairwise non-proportional) linear forms in $x_1, x_2, x_3, x_4$ for $i = 1, \cdots, 6$?

If the answer is no, is there a somewhat straightforward proof as to why (this must be a geometric reason, since the above forms are all allowed to have coefficients in an algebraically closed field)?

If the answer is yes, can one choose $A,B$ to have coefficients in $\mathbb{Q}$ instead? Ideally, the polynomial $D$ will then be irreducible over $\mathbb{Q}$ (but still splits over $\overline{\mathbb{Q}}$).

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    $\begingroup$ Well, there is a stupid answer, just pick a linear form $L$ and put $A = L^2$, $B = L^3$, but this is likely not what you want. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 7 at 19:40
  • $\begingroup$ @AlekseiKulikov indeed, I do not want that degenerate case. I edited the question to reflect this $\endgroup$
    – Stanley Yao Xiao
    Commented Dec 7 at 19:53
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    $\begingroup$ Well, another option is to pick literally any forms $A, B$ assuming they only depend on $x_1$ and $x_2$, then everything will factor again over $\mathbb{C}$ but not necessarily over $\mathbb{Q}$. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 7 at 20:17
  • $\begingroup$ If an elegant geometric argument is hard to find, there is also a brutal approach of looking at the Brill equations for $D$ with generic $A$ and $B$, with the help of a computer algebra system. See mathoverflow.net/a/109340/7410 $\endgroup$
    – Abdelmalek Abdesselam
    Commented Dec 7 at 20:27

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