= Solution
The <traction> on the oriented solid element is
$$
\boxed{\boldsymbol\sigma=\tau\mathbf n,\qquad\sigma_i=\tau_{ij}n_j.}
$$
With constant Lamé coefficients and derivatives in the reference variables,
$$
\partial_j\tau_{ij}
=\lambda\partial_i(\partial_ku_k)
+\mu\partial_j(\partial_ju_i+\partial_iu_j)
=(\lambda+\mu)\partial_i(\nabla\cdot\mathbf u)+\mu\nabla^2u_i.
$$
Substitution into <balance of linear momentum> gives
$$
\boxed{\rho\mathbf u_{tt}=(\lambda+\mu)\nabla(\nabla\cdot\mathbf u)
+\mu\nabla^2\mathbf u+\rho\mathbf g.}
$$
To check the <vector> identity componentwise, contract the two alternating <tensors>:
$$
[\nabla\times(\nabla\times\mathbf u)]_i
=\epsilon_{ijk}\epsilon_{klm}\partial_j\partial_lu_m
=\partial_i\partial_ju_j-\partial_j\partial_ju_i.
$$
Rearranging proves $\nabla^2\mathbf u=\nabla(\nabla\cdot\mathbf u)-\nabla\times(\nabla\times\mathbf u)$. Therefore
$$
\boxed{\rho\mathbf u_{tt}=(\lambda+2\mu)\nabla(\nabla\cdot\mathbf u)
-\mu\nabla\times(\nabla\times\mathbf u)+\rho\mathbf g.}
$$
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