= Solution
{c}
For a <hyperelastic material> $T_R(F)=DW(F)$. <Material frame indifference> of this first Piola-Kirchhoff law means
$$
\boxed{T_R(QF)=Q\,T_R(F)\quad(Q\in SO(3)).}
$$
The spatial <force> components rotate under a superposed rigid <rotation>, whereas the reference <normal> does not.
If $W(QF)=W(F)$, differentiate with respect to $F$ in direction $H$ to obtain $DW(QF):(QH)=DW(F):H$, hence the displayed <stress> transformation. Conversely, assume that <stress> transformation. For fixed $Q$, the <derivative> of $W(QF)-W(F)$ with respect to $F$ vanishes. The positive-determinant <matrix> domain is <connected>, as follows from <polar decomposition> and the <connectedness> of $SO(3)$ and of the <positive-definite> <cone>. Thus $W(QF)-W(F)=c(Q)$, independent of $F$. Iterating gives
$$
W(Q^nF)=W(F)+n c(Q)\qquad(n\in\mathbb Z).
$$
The assumed lower bound on $W$ forces $c(Q)=0$: either nonzero sign would tend to minus infinity for one direction of integer $n$. This proves the required equivalence and explains the use of the lower-bound hypothesis.
In the frame-indifferent setting, <isotropy> means invariance under reference <rotations> as well: $W(FQ)=W(F)$ for $Q\in SO(3)$. The distinction matters: left <rotations> encode changes of observer, and right <rotations> encode material symmetry.
An orientation-preserving <singular value decomposition> is $F=R\operatorname{diag}(v_1,v_2,v_3)S$, with $R,S\in SO(3)$ and $v_i>0$. <Material frame indifference> and <isotropy> remove $R,S$, leaving a function $\Phi$ of the stretches. Permuting the stretches using signed permutation <rotations> does not change the value, so $\Phi$ is <symmetric>. Conversely a <symmetric> stretch function is unchanged by both left and right <rotations>, in particular is <isotropic>.
The <eigenvalues> of $B=FF^T$ are $v_i^2$. Its three principal invariants are
$$
I=\operatorname{tr}B=\sum_iv_i^2,\quad
II=\tfrac12[(\operatorname{tr}B)^2-\operatorname{tr}(B^2)]
=\sum_{i<j}v_i^2v_j^2,\quad III=\det B=\prod_iv_i^2.
$$
They determine the unordered positive <eigenvalues> as the roots of $\lambda^3-I\lambda^2+II\lambda-III=0$. Therefore a <symmetric> stretch function defines a well-defined $h(I,II,III)$ on the admissible invariant triples. Conversely a function of these invariants is a <symmetric> stretch function, proving the equivalence of all three descriptions.
The printed domain of $\Phi$ must be $(0,\infty)^3$, since it takes three arguments. Also <isotropy> alone as right invariance would not imply the spectral representation without <material frame indifference>: $\operatorname{tr}(A FF^T)$ for a fixed nonscalar <symmetric> <positive-definite> $A$ is right invariant but not generally objective. The equivalence here uses the frame-indifferent constitutive setting.
For the specific energy, <Newton identities> for the three <eigenvalues> gives $\sum_i v_i^6=I^3-3I\,II+3III$, while $v_1v_2v_3=J=\sqrt{III}$. Thus
$$
\boxed{W=\alpha I+\beta(I^3-3I\,II+3III)+\gamma III^{-1/2}.}
$$
To differentiate it, use
$$
D_FI=2F,\qquad D_FII=2(IF-BF),\qquad
D_FIII=2J\operatorname{cof}F.
$$
These follow from $I=|F|^2$, the displayed <trace> formula for $II$, and $III=J^2$. The three partial derivatives of $h$ are $h_I=\alpha+3\beta I^2-3\beta II$, $h_{II}=-3\beta I$ and $h_{III}=3\beta-\gamma III^{-3/2}/2$. The <chain rule> therefore gives
$$
T_R=(2\alpha-6\beta II)F+6\beta I\,BF
+(6\beta J-\gamma J^{-2})\operatorname{cof}F.
$$
Finally $-6\beta II=-3\beta(\operatorname{tr}B)^2+3\beta\operatorname{tr}(B^2)$, and $BF=FF^TF$. This is precisely the requested expression, with the coefficient of $F$ separated from the $BF$ and <cofactor> terms. The positive sign of the $6\beta J$ term comes from the $3\beta III$ contribution in the energy.
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