= Solution
The six critical <Fourier modes> are at $\pm\mathbf k_1,\pm\mathbf k_2,\pm\mathbf k_3$, with $\mathbf k_1+\mathbf k_2+\mathbf k_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the <extended centre manifold> parametrization is consequently
$$
u(\mathbf x,t)=\sum_{i=1}^3 A_i(t)e^{i\mathbf k_i\cdot\mathbf x}+\text{complex conjugate}+\text{slaved higher harmonics}.
$$
The displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full <centre manifold> need not vanish identically.
A translation $\mathbf d$ acts by $A_i\mapsto e^{i\mathbf k_i\cdot\mathbf d}A_i$. Each <monomial> in the equation for $A_i$ must therefore carry wavevector $\mathbf k_i$. Since $-\mathbf k_{i+1}-\mathbf k_{i+2}=\mathbf k_i$, the allowed quadratic term is $A_{i+1}^*A_{i+2}^*$. At cubic order the allowed terms are $A_i|A_i|^2,A_i|A_{i+1}|^2,A_i|A_{i+2}|^2$. Rotation through $60$ degrees acts, with one choice of orientation, as
$$
(A_1,A_2,A_3)\longmapsto(A_2^*,A_3^*,A_1^*).
$$
Its square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; <equivariance> under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the <chiral hexagonal amplitude equations> are
$$
\boxed{\dot A_i=\mu A_i+\alpha A_{i+1}^*A_{i+2}^*
-A_i(a|A_i|^2+b|A_{i+1}|^2+c|A_{i+2}|^2),}
$$
at the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase <equivariance> requires $|A_{i+2}|^2$.
Write $A_i=r_ie^{i\phi_i}$, with all three amplitudes nonzero, and let $\Phi=\phi_1+\phi_2+\phi_3$. Separating real and imaginary parts gives
$$
\dot r_i=\mu r_i+\alpha r_{i+1}r_{i+2}\cos\Phi
-r_i(ar_i^2+br_{i+1}^2+cr_{i+2}^2),\qquad
\dot\phi_i=-\alpha\frac{r_{i+1}r_{i+2}}{r_i}\sin\Phi.
$$
Hence the <phase locking of a resonant hexagonal triad> obeys
$$
\dot\Phi=-\alpha\left(\frac{r_2r_3}{r_1}+\frac{r_3r_1}{r_2}+\frac{r_1r_2}{r_3}\right)\sin\Phi.
$$
For $\alpha>0$, $\Phi=0$ is phase stable and $\Phi=\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus \b[a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin]. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu<0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.
In the real nonnegative subspace, a roll is
$$
\boxed{(A_1,A_2,A_3)=(r,0,0),\qquad r^2=\mu/a,\qquad\mu>0,}
$$
or a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu+\alpha r-sr^2=0$, giving
$$
\boxed{r_\pm=\frac{\alpha\pm\sqrt{\alpha^2+4s\mu}}{2s},}
$$
with only positive roots retained. The upper branch exists for $\mu\geq-\alpha^2/(4s)$; the lower branch is positive for $-\alpha^2/(4s)\leq\mu<0$. They meet at a fold.
For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus \b[there are no nontrivial rectangles].
In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the <uniqueness of positive hexagon equilibria with straddling cross-couplings> argument. Put $x_i=r_i^2$, $p=r_1r_2r_3>0$. The equilibrium equations imply
$$
a x_i+b x_{i+1}+c x_{i+2}-\frac{\alpha p}{x_i}=\mu.
$$
Consider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\geq x_3$, subtracting equations one and two gives
$$
B(x_2-x_3)+C(x_3-x_1)=\alpha p(1/x_1-1/x_2).
$$
The left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\leq x_3$, subtract equations two and three instead:
$$
B(x_3-x_1)+C(x_1-x_2)=\alpha p(1/x_2-1/x_3).
$$
Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.
At a roll the active radial <eigenvalue> is $-2\mu$. The two inactive amplitudes have real-part <linearization>
$$
\begin{pmatrix}\mu(1-c/a)&\alpha r\\\alpha r&\mu(1-b/a)\end{pmatrix}.
$$
Its determinant is $\mu^2(a-c)(a-b)/a^2-\alpha^2\mu/a<0$. It therefore has a positive <eigenvalue>: \b[the rolls are unstable for every $\mu>0$], which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.
At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2ar^2-\alpha r$ and off-diagonals $p=\alpha r-2br^2$, $q=\alpha r-2cr^2$. Its uniform <eigenvector> $(1,1,1)$ has <eigenvalue>
$$
\lambda_{\parallel}=d+p+q=\alpha r-2sr^2.
$$
The other two <eigenvectors> lie in the zero-sum subspace; using the two nontrivial cube roots of unity gives
$$
\boxed{\lambda_\pm=(b+c-2a)r^2-2\alpha r\ \pm i\sqrt3(b-c)r^2.}
$$
For sufficiently large $\mu$, $r\sim\sqrt{\mu/s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus \b[the hexagons are unstable at large $\mu$], while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has <eigenvalue> $-3\alpha r$, so the amplitude instability is genuine.
More precisely, the upper branch is amplitude stable between its fold and the <chiral hexagon Hopf threshold>
$$
r_H=\frac{2\alpha}{b+c-2a},\qquad
\mu_H=\frac{2\alpha^2(4a+b+c)}{(b+c-2a)^2}.
$$
At this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a <Hopf bifurcation>. The linear calculation locates the crossing; the nonlinear <normal form> decides the stability and direction of its periodic branch.
The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,
$$
\dot S=2\mu S+6\alpha r_1r_2r_3-2aS^2-2(b+c-2a)\sum_{i<j}r_i^2r_j^2.
$$
Since $r_1r_2r_3\leq(S/3)^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the <Küppers–Lortz instability>. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha=0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha>0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.
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