Let $\underline{u}$ be a $C^4(\mathbb{R^3})$ (vector-valued) solution of the "displacement equation of equilibrium"
$$ \mu \, \Delta\underline{u} + (\lambda+\mu)\,\nabla \operatorname{Div} (\underline{u}) = 0 $$ where $\mu$ and $\lambda$ are (usually positive) constants. How can I prove that $\operatorname{Div}(\underline{u})$ is harmonic? I have tried to consider the divergence of the above equation, thus obtaining $$ \mu \operatorname{Div}(\Delta\underline{u}) + (\lambda+\mu)\Delta( \operatorname{Div} (\underline{u})) = 0 $$ but I can't figure out why one could claim that $\operatorname{Div}(\Delta\underline{u})=0$...
For functions from $C^4$ the Laplacian and the divergence commute (the order of differentiation for third derivatives does not matter), so we have $$ \mu \operatorname{Div}(\Delta\underline{u}) + (\lambda+\mu)\Delta( \operatorname{Div} (\underline{u})) = \mu \Delta(\operatorname{Div}(\underline{u})) + (\lambda+\mu)\Delta( \operatorname{Div} (\underline{u}))=(\lambda+2\mu)\Delta( \operatorname{Div} (\underline{u}))=0. $$