= Solution
Write $B_{i\alpha}=\dot A_{i\alpha}$ and assume that the material constraint is regular, so $F_{,A}\ne0$. Differentiating the constraint shows that admissible rates of the <deformation gradient> satisfy $F_{,A_{i\alpha}}B_{i\alpha}=0$. The given stress-power identity says that $W_{,A}-N^T$ annihilates this tangent hyperplane. Its orthogonal complement is the one-dimensional span of $F_{,A}$, so a <Lagrange multiplier> gives the <constraint reaction in hyperelastic stress>:
$$
\boxed{N_{\alpha i}=W_{,A_{i\alpha}}+qF_{,A_{i\alpha}}.}
$$
The sign of $q$ is a convention; the <virtual work> identity determines no constitutive value for this reaction. Regularity matters: a singular equation describing the same constraint need not have a nonzero gradient.
For <incompressibility>, take $F=\det A-1$. Differentiating the expression using the <Levi-Civita symbol> gives
$$
\frac{\partial\det A}{\partial A_{i\alpha}}=\frac12\epsilon_{\alpha\beta\gamma}\epsilon_{ijk}A_{j\beta}A_{k\gamma}=(\det A)(A^{-1})_{\alpha i}.
$$
The middle expression is the corresponding entry of the <cofactor matrix>. Since $\det A=1$, the <nominal stress> is therefore
$$
\boxed{N_{\alpha i}=W_{,A_{i\alpha}}+q(A^{-1})_{\alpha i}.}
$$
For the <neo-Hookean solid>, $W_{,A_{i\alpha}}=\mu A_{i\alpha}$, so in the material-first convention $N=\mu A^T+qA^{-1}$. The <identity matrix> satisfies the prescribed <dead loading> with $q=T-\mu$, proving that the undeformed cube is always an equilibrium.
To obtain the <homogeneous tensile bifurcation of a neo-Hookean cube>, first seek $A=\operatorname{diag}(\lambda,\lambda,\lambda^{-2})$, with $\lambda>0$. This has unit <determinant>. The two distinct diagonal stress equations are $\mu\lambda+q/\lambda=T$ and $\mu\lambda^{-2}+q\lambda^2=T$. Eliminating $q$ gives
$$
(\lambda^3-1)\left[\mu(\lambda+\lambda^{-2})-T\right]=0.
$$
Thus the nontrivial branch is
$$
\boxed{t=\lambda+\lambda^{-2},\qquad \lambda^3-t\lambda^2+1=0,\qquad q=\mu/\lambda,\quad t=T/\mu.}
$$
The function $h(\lambda)=\lambda+\lambda^{-2}$ decreases on $(0,2^{1/3})$ and increases on $(2^{1/3},\infty)$, because $h'=1-2\lambda^{-3}$. Its minimum is $h(2^{1/3})=3/2^{2/3}$. Above this threshold there are two positive roots $\lambda_-<2^{1/3}<\lambda_+$. For $t\ne2$, neither root is one, and the three choices of the exceptional coordinate for each root give six distinct diagonal <deformation gradients>, in addition to $I$.
At $t=2$, however, $\lambda_-=1$ and the three smaller-root matrices coincide with $I$: it would be incorrect to count those as three new deformations. The requested existence of seven distinct homogeneous deformations nevertheless holds even at this value. For any unit vector $n$, put
$$
A(n)=\lambda_+ I+(\lambda_+^{-2}-\lambda_+)n n^T.
$$
Its positive <eigenvalues> are $\lambda_+,\lambda_+,\lambda_+^{-2}$, and its <determinant> is one. Since $A(n)$ is an orthogonal conjugate of the diagonal solution, the same $q=\mu/\lambda_+$ gives $\mu A(n)^T+qA(n)^{-1}=T I$. Choose the six directions
$$
e_1,\ e_2,\ e_3,\ \frac{e_1+e_2}{\sqrt2},\ \frac{e_1+e_3}{\sqrt2},\ \frac{e_2+e_3}{\sqrt2}.
$$
Their <outer products> are distinct, and $\lambda_+>2^{1/3}>1$, so the six $A(n)$ are distinct from one another and from $I$. Constant <nominal stress> satisfies equilibrium and the prescribed surface <tractions>. Hence \b[seven distinct homogeneous equilibria exist whenever $T/\mu>3/2^{2/3}$]. This is an existence count, not an exhaustive classification: rotated principal axes actually give a continuous family. The conventional seven axis-aligned positive-stretch branches coalesce to four distinct matrices at $T/\mu=2$.
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