= Solution
Contract the given <isotropic fourth-rank tensor> with the <strain> array. The <stress tensor> is
$$
P_{ij}=\lambda\delta_{ij}p_{kk}+\mu p_{ij}+\nu p_{ji}.
$$
Write $p_{ij}=p\delta_{ij}+s_{ij}+\epsilon_{ijk}a_k$, where $s_{ij}$ is symmetric and traceless. Then $p_{kk}=3p$ and $p_{ji}=p\delta_{ij}+s_{ij}-\epsilon_{ijk}a_k$. Substitution gives
$$
P_{ij}=(3\lambda+\mu+\nu)p\delta_{ij}+(\mu+\nu)s_{ij}+(\mu-\nu)\epsilon_{ijk}a_k.
$$
By uniqueness of the <scalar, symmetric-traceless and axial tensor decomposition>,
$$
\boxed{P=(3\lambda+\mu+\nu)p,\qquad S_{ij}=(\mu+\nu)s_{ij},\qquad A_k=(\mu-\nu)a_k.}
$$
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.
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