 Solution

Contract the given isotropic fourth-rank tensor with the strain array. The stress tensor isWrite $p_{ij}=p{\delta }_{ij}+s_{ij}+{ϵ}_{ijk}{a}_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p{\delta }_{ij}+s_{ij}-{ϵ}_{ijk}{a}_k$. Substitution givesBy uniqueness of the scalar, symmetric-traceless and axial tensor decomposition,This is isotropic elasticity on tensor components: the three tensor sectors do not mix. If the strain is symmetric, $a_k=0$, hence $A_k=0$ and $\boxed{P_{ij}=P_{ji}}$. No equality between $\mu$ and $\nu$ is needed for that conclusion.