Q:

Vector identity proof

I am told that $\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\triangle A$.
I'm then asked to use this to show that $\nabla\times(\nabla\times(\nabla\times A))=-\triangle(\nabla\times A).$
So \begin{align*}\nabla\times(\nabla\times(\nabla\times A))
&=\nabla\times(\nabla(\nabla\cdot A)-\triangle A)\\
&=\nabla\times(\nabla(\nabla\cdot A)-\nabla\cdot\nabla A))\\
&=\epsilon_{ijk}\partial_j(\partial_k\partial_mA_m-\partial_k\partial_kA)
\end{align*}
It doesn't look like this is taking me anywhere though. What should I be doing?

A:

You should keep the inner $\nabla \times A$ term and calculate from the 'outside'. Then using the stated identity:
$$ \nabla\times (\nabla \times (\nabla\times A)) =\nabla (\nabla \cdot (\nabla\times A)) - \Delta (\nabla \times A) =- \Delta (\nabla \times A)  $$
(in the first term one has $\nabla \cdot (\nabla\times A)=0$ ).

