Showing that $\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$
Solution 1:
For some details missing from the derivation below, see my answer to this question.
First, we write the LHS in terms of its components, using the Kronecker delta and Levi-Civita symbols. (Note: I'll drop the vector arrow on $\vec{A}$ but it's a vector)
$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\partial_{j}\,(\nabla \times A)_{k} = \epsilon_{ijk}\,\partial_{j}\,(\epsilon_{krs}\,\partial_{r}\,A_{s}) $$
The $\epsilon_{krs}$ are constants, so
$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\epsilon_{krs}\,\partial_{j}\,\partial_{r}\,A_{s} $$
But
$$\epsilon_{ijk}\,\epsilon_{krs} = \delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr} $$
so
$$[\,\nabla \times (\nabla \times A)\,]_{i} = (\delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr})\, \partial_{j}\,\partial_{r}\,A_{s} = \delta_{ir}\,\delta_{js}\,\partial_{j}\,\partial_{r}\,A_{s} - \delta_{is}\,\delta_{jr}\,\partial_{j}\,\partial_{r}\,A_{s} $$
Simplifying,
$$[\,\nabla \times (\nabla \times A)\,]_{i} = \partial_{j}\,\partial_{i}\,A_{j} - \partial_{j}\,\partial_{j}\,A_{i} = \partial_{i}\,(\partial_{j}\,A_{j}) - (\partial_{j}\,\partial_{j})\,A_{i} = \partial_{i}\,(\nabla \cdot A) - \nabla^2A_i $$
so
$$\nabla \times (\nabla \times A) = \nabla(\nabla \cdot A) - \nabla^2A $$
Solution 2:
All you need to know is the definitions of (just taking $\mathbb{R}^3$ for simplicity) $$ \nabla=\left[\matrix{\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}}\right]\quad\text{(a symbolic vector)},\qquad \vec A=\left[\matrix{A_1(x,y,z)\\A_2(x,y,z)\\A_3(x,y,z)}\right]\qquad\text{(a vector)} $$ the scalar product and the vector product of two vectors. Well, also $\Delta=\nabla\cdot\nabla$. Then just apply the products formally to the vector coordinates and keep the order so that the derivatives affect functions.
Solution 3:
Similar to @wltrup with notational changes, we have
$$\begin{align} \nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\ &=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\ &=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\ &=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\ &=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\ &=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5 \end{align}$$
In going from $(1)$ to $(2)$ we made use of the vector triple product. Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.
In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.
In going from $(3)$ to $(4)$, we rearranged terms.
In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.
Solution 4:
I am coming to this party about three years too late, but the answers posted here, and some here as well: enter link description here, may leave readers with the false impression that the vector Laplacian acting on a vector field (appearing on the right-hand side) always takes the form $$\begin{align} \nabla^2 \vec{A} = \nabla^2(A_i)\,\hat{e}_i ~~~ {\text{(summation convention in force)}} \tag 1 \end{align}$$ in an arbitrary basis and coordinate system of $\mathbb{R}^3$, where $\nabla^2$ is the Laplacian operator in the coordinates associated with the basis. This is true only in a cartesian coordinate system and its orthonormal basis. For example, in the cylindrical coordinate system $\{\rho, \phi, z\}$, with orthogonal (but not orthonormal) basis $\{\hat{e}_\rho, \hat{e}_\phi. \hat{e}_z\}$, it might be supposed from these derivations that $$\begin{align} \nabla^2 \vec{A} = \nabla^2(A_\rho)\,\hat{e}_\rho + \nabla^2(A_\phi)\,\hat{e}_\phi + \nabla^2(A_z)\,\hat{e}_z, \end{align}$$ where $$\begin{align} \nabla^2(f) = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2} \end{align}$$ is the Laplacian operator in cylindrical coordinates acting on a scalar function $f$. This is quite false, as I first learned long ago from the textbook ``Mechanics of Deformable Bodies'', by the master mathematical physicist Arnold Sommerfeld. In the answers to Problems I.3 and I.4 of the book, he lays bare the fallacy of this assumption. The first two components of $\nabla^2{\vec{A}}$ contain extra terms besides $\nabla^2(A_\rho)$ and $\nabla^2(A_\phi)$.
In fact, Sommerfeld clearly states in Chapter I, page 23, what seems to have been overlooked, or at any rate, not emphasized, in many of the answers posted here: the formula $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla\cdot \vec{A}) - \nabla^2\vec{A}$ is an identity in the sense of (1) being true, only in cartesian coordinates. For any other coordinate system and basis, the formula serves to define $\nabla^2\vec{A}$ by $$\begin{align} \nabla^2\vec{A} := \nabla(\nabla\cdot \vec{A}) - \nabla \times (\nabla \times \vec{A}), \end{align}$$ each term on the right-hand side being a well-defined operation in any coordinate system. So the upshot is that the formula can be shown to be true only in a cartesian coordinate system (as was done in the answers here); in any other coordinate system the formula instead defines the vector Laplacian acting on a vector field.