How do I prove this vector calculus relationship? If $\nabla^2\phi = 0$ and $\vec m$ is a constant vector, show that 
$$\nabla \times (\vec m \times \nabla \phi) + \nabla (\vec m \cdot \nabla \phi) = 0$$
I tried to expand the following, which gives me 
$$(\nabla \times \vec m) \times(\nabla \times \nabla \phi) + \nabla \vec m \cdot \nabla^2 \phi = 0$$
The second term is zero, but how do I make the first term zero as well?
 A: I think your expansions might not be quite correct. Following this answer here, we use the Levi-Civita symbol $\varepsilon_{ijk}$ with the following identities:
\begin{align*}
\varepsilon_{ijk}&=\varepsilon_{jki}=\varepsilon_{kij} \\
\varepsilon_{ijk}\,\varepsilon_{ilm}&=\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl}\\
(F\times G)_i&=\varepsilon_{ijk}F_jG_k.
\end{align*}
Here, we're using the Einstein summation convention all over the place. So, let's see what we have:
\begin{align*}
[\nabla\times(\vec{m}\times\nabla\phi)+\nabla(\vec{m}\cdot\nabla\phi)]_i&=\varepsilon_{ijk}\,\partial_j\,(\vec{m}\times\nabla\phi)_k+\partial_i\,m_n\,\partial_n\,\phi \\
&=\varepsilon_{ijk}\,\partial_j\,\varepsilon_{klm}\,m_l\,(\nabla\phi)_m+\partial_i\,m_n\,\partial_n\,\phi \\
&=\varepsilon_{ijk}\,\varepsilon_{klm}\,m_l\,\partial_j\,\partial_m\,\phi+m_n\,\partial_i\,\partial_n\,\phi \\
&=\varepsilon_{kij}\,\varepsilon_{klm}\,m_l\,\partial_j\,\partial_m\,\phi+m_n\,\partial_i\,\partial_n\,\phi \\
&=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})\,m_l\,\partial_j\,\partial_m\,\phi+m_n\,\partial_i\,\partial_n\,\phi \\
&=\delta_{il}\delta_{jm}\,m_l\,\partial_j\,\partial_m\,\phi-\delta_{im}\delta_{jl}\,m_l\,\partial_j\,\partial_m\,\phi+m_n\,\partial_i\,\partial_n\,\phi \\
&=m_i\,\partial_m\,\partial_m\,\phi-m_j\,\partial_j\,\partial_i\,\phi+m_n\,\partial_i\,\partial_n\,\phi \\
&=[m_i\,\partial_m\,\partial_m-m_j\,\partial_j\,\partial_i+m_n\,\partial_i\,\partial_n]\,\phi.
\end{align*}
Here's where we can see how we get zero out of it all. We note that any repeated indices here are summed over, and thus the summed indices are dummy. We rewrite using the same ones, therefore, to obtain
$$[\nabla\times(\vec{m}\times\nabla\phi)+\nabla(\vec{m}\cdot\nabla\phi)]_i=
[m_i\,\partial_j\,\partial_j-m_j\,\partial_j\,\partial_i+m_j\,\partial_i\,\partial_j]\,\phi.$$
By Clairaut's Theorem, $\partial_i\,\partial_j\,\phi=\partial_j\,\partial_i\,\phi,$ so the second two terms vanish, leaving
$$[\nabla\times(\vec{m}\times\nabla\phi)+\nabla(\vec{m}\cdot\nabla\phi)]_i=
m_i\,\partial_j\,\partial_j\,\phi.$$
But, because we were given that $\nabla^2\phi=0$, this last expression, also, is zero. QED.
