= Solution
{c}
Take two time derivatives at fixed reference position $\mathbf X$:
$$
\ddot{\mathbf x}=\ddot{\mathbf F}\mathbf X+\ddot{\mathbf c}.
$$
The constraint $\det\mathbf F=1$ ensures that the <deformation gradient> is invertible, so $\mathbf X=\mathbf F^{-1}(\mathbf x-\mathbf c)$. Thus the spatial <material acceleration> is the affine <vector field>
$$
\ddot{\mathbf x}=\mathbf A(t)(\mathbf x-\mathbf c(t))+\ddot{\mathbf c}(t),
\qquad \mathbf A=\ddot{\mathbf F}\mathbf F^{-1}.
$$
Its spatial <gradient> is $\mathbf A$, and its <curl> is
$$
\nabla\times\ddot{\mathbf x}
=(A_{32}-A_{23},\ A_{13}-A_{31},\ A_{21}-A_{12}).
$$
Part (b)(iii) therefore requires
$$
\boxed{\ddot{\mathbf F}\mathbf F^{-1}
=\left(\ddot{\mathbf F}\mathbf F^{-1}\right)^T.}
$$
Equivalently, multiplying on the left by $\mathbf F^T$ and on the right by $\mathbf F$ gives
$$
\boxed{\mathbf F^T\ddot{\mathbf F}=\ddot{\mathbf F}^{\,T}\mathbf F.}
$$
The translation $\mathbf c(t)$ contributes no spatial <curl> and is unrestricted by this condition. A useful equivalent first integral is
$$
\frac d{dt}\left(\mathbf F^T\dot{\mathbf F}-\dot{\mathbf F}^{\,T}\mathbf F\right)=\mathbf0:
$$
the antisymmetric <matrix> in parentheses must be constant, and need not vanish. The separate constraint $\det\mathbf F=1$ is still required; differentiating it also gives $\operatorname{tr}(\dot{\mathbf F}\mathbf F^{-1})=0$.
The condition is understood in the setting of part (b), where a static family already exists under the prescribed <body force>. For a homogeneous elastic <deformation gradient>, the elastic <stress> $\mathbf S(\mathbf F)$ is spatially constant. Static <balance of linear momentum> consequently gives $\nabla p_0=\rho\mathbf b$, ensuring that $\mathbf b$ has a <scalar potential>. If the static-family assumption were dropped and an arbitrary time-independent <body force> allowed, the general necessary condition would instead be $\nabla\times\ddot{\mathbf x}=\nabla\times\mathbf b$.
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