Expired by this question Show determinant of matrix is non-zero I am moved to ask:
Given integers $a,b,c,$ and cubic form $$ f(a,b,c) = a^3+2 b^3-6 a b c+4 c^3 = \left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right|, $$ what primes $p$ can be integrally represented as $$ p = f(a,b,c)? $$
I think it is $3,$ all primes $p \equiv 2 \pmod 3,$ and all $p = u^2 + 27 v^2$ in integers, but not any $q = 4 u^2 + 2 u v + 7 v^2.$ I checked for $p < 10000.$
Note that, if $-p$ is represented, so is $p.$
Although it does not finish things, note that if $f$ integrally represents both $m,n$ then it represents $mn.$ That is because $f(a,b,c) = \det(aI + b X + c X^2),$ where $$ X = \begin{bmatrix} 0 & 0 & 2\\1 & 0 & 0\\ 0 & 1 & 0\end{bmatrix}. $$ Then $X^3 = 2 I$ and $X^4 = 2 X.$
I once asked a guy at MSRI about pretty much the same problem, only instead of the important polynomial being $\lambda^3 - 2$ it was $\lambda^3 - \lambda^2 - \lambda - 1.$ The phrase norm forms came up, and he laughed at me.
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p a b c
2 0 1 0
3 -1 0 1
5 1 0 1
11 -1 1 1
17 -1 -1 2
23 1 1 2
29 -3 10 -6
31 -1 26 -20
41 1 -2 2
43 1 -1 2
47 -1 4 -2
53 1 -4 3
59 1 3 -2
71 -1 2 2
83 3 1 3
89 1 2 3
101 3 -7 4
107 -1 0 3
109 1 -12 9
113 1 4 2
127 -1 16 -12
131 3 3 -1
137 -3 1 3
149 1 4 -1
157 -1 5 -2
167 -3 3 2
173 -3 7 -3
179 1 -31 24
191 -1 -2 4
197 5 2 -1
223 1 5 2
227 3 -2 3
229 -1 -1 4
233 1 5 -3
239 1 3 4
251 -1 -4 5
257 1 0 4
263 3 4 -3
269 -1 9 -6
277 1 5 -1
281 -1 1 4
283 -1 8 -5
293 1 -9 7
307 -1 4 3
311 3 3 5
317 -3 5 1
347 3 -12 8
353 3 -1 4
359 -5 23 -15
383 -5 28 -19
389 -3 2 4
397 1 7 -5
401 1 -5 5
419 3 6 5
431 1 -7 6
433 -1 -5 6
439 3 1 5
443 3 -4 4
449 1 8 -6
457 1 2 5
461 5 4 -2
467 -1 -1 5
479 -1 4 4
491 3 18 -16
499 -1 0 5
503 5 3 6
509 1 4 5
521 5 5 -1
557 -1 89 -70
563 3 6 -1
569 -1 7 -2
587 3 4 6
593 -7 2 5
599 1 7 -4
601 1 -22 17
617 -5 -59 50
641 3 23 -20
643 -1 3 5
647 -1 14 -10
653 1 16 -13
659 3 -10 7
677 -1 -11 10
683 -5 5 3
691 3 -2 5
701 -1 -3 6
719 5 5 -4
727 3 -5 5
733 3 9 -8
739 1 7 -2
743 3 -3 5
761 1 -14 11
773 5 -1 5
797 -3 3 5
809 1 2 6
811 1 3 6
821 3 7 6
827 -1 11 -7
839 -3 5 4
857 -9 5 4
863 -1 0 6
881 1 -4 6
887 7 3 7
911 -1 -5 7
919 -1 7 3
929 9 2 -2
941 9 3 -1
947 -3 1 6
953 -7 26 -16
971 -1 8 -1
977 1 7 5
983 3 7 -3
997 3 -11 8
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The discriminant of $\Bbb Q \subset \Bbb Q(\sqrt[3]2)$ is $-108$, and the Minkowski bound for this extension is $\frac {3!}{3^3} \frac 4 \pi \sqrt {108} \approx 2.940$. So to prove that this number field has class number $1$ we only need to find a way to represent $2$, and $2$ is indeed represented by $(0,1,0)$. Thus $p$ is represented by this norm form if and only if the ideal $(p)$ has an ideal factor of norm $p$, which happens if and only if $2$ is a cube modulo $p$.
If $p \equiv 2 \pmod 3$ then any nonzero element of $\Bbb F_p$ has one cube root in $\Bbb F_p$ and two cube roots in $\Bbb F_p^2$, so $2$ is a cube modulo $p$.
If $p \equiv 1 \pmod 3$ then $(p)$ splits in $\Bbb Q(\zeta_3)$, and $2$ is a cube if and only it further splits in $\Bbb Q(\zeta_3,\sqrt[3]2)$. Since $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$ is an abelian extension, it is a ray class field for some modulus $\mathfrak m$ of $\Bbb Q(\zeta_3)$.
Working modulo $6$, we have $(a+b\zeta_3)^3 = (a^3+b^3) - 3ab^2+3ab(a-b)\zeta_3 = a^3 + b^3-3ab^2 \in \Bbb Z/6 \Bbb Z$, and thus for any $a,b,c \in \Bbb Z[\zeta_3]$, $a^3+2b^3+4c^3-6abc = a^3+2b^3+4c^3 \in \Bbb Z/6\Bbb Z$. So, norms that are coprime to $6$ are units ($\pm 1$) modulo $6$. So $\Bbb Q(\zeta_3,\sqrt[3]2)$ is an extension of the ray class field of modulus $(6)$ for $\Bbb Q(\zeta_3)$.
On the other hand, $G = (\Bbb Z[\zeta_3]/(6))^*/\langle \overline{\zeta_6} \rangle$ is isomorphic to $\Bbb Z/3 \Bbb Z$, which is the Galois group of the extension $\Bbb Q(\zeta_3) \subset \Bbb Q(\zeta_3,\sqrt[3]2)$, so $\mathfrak m = (6)$, and $2$ is a cube modulo $p$ if and onlt if $p \equiv 2 \mod 3$ or $p = a^2-ab+b^2$ where $a+\zeta_3 b$ is congruent modulo $6$ to one of $\{1,1+\zeta_3,\zeta_3,-1,-1- \zeta_3,- \zeta_3\}$.
Each element of $G$ (modulo complex conjugation) corresponds to a class of primitive binary quadratic forms of discriminant $-108$, or a corresponding lattice class (modulo multiplication by a unit and complex conjugation) whose endomorphism ring is $\Bbb Z[3\sqrt{-3}]$:
$\Lambda = \langle 1, 3\sqrt{-3} \rangle$ is a lattice corresponding to the neutral element of $G$ : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the neutral class.
while $\Lambda = \langle 2, \frac {1+3\sqrt{-3}}2 \rangle$ corresponds to the other two classes : it contains $(6)$ and the numbers coprime with $(6)$ it meets all fall in the other two classes.
So if $p \equiv 1 \pmod 3$, then $p$ is represented either as $a^2 + 27b^2$ (when $2$ is a cube) or as $4u^2 \pm 2uv + 7v^2$ (when $2$ is not a cube), and never both at the same time.