Identities with Div, Grad, Curl In physics there are lots of identities like:
$$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - (\nabla \cdot \nabla) A$$
I'm wondering if there is an algorithmic algebraic method to prove and/or derive these identities (something like using $e^{i\theta}$ to prove trigonometric identities)?
 A: The most straightforward route I know of is through Einstein tensor notation, which renders the derivation of such identities largely mechanical by virtue of suppressing distracting notational noise. (I should warn you that I'm largely self-taught on this subject, so the following may contain errors of rigour, but it "works" as long as you're working in Cartesian coordinates with the Euclidean metric.)
Let me work through your example to illustrate. In tensor notation, the $i$th component of the curl of a vector field $v_i$ is given by $\varepsilon_{ijk}\partial_j v_k$, where $\varepsilon_{ijk}$ is the Levi-Civita symbol, that takes the values $1$, $-1$ and $0$ depending on the order in which the coordinates appear in the subscripts $ijk$. You can think of this as a notational shorthand for the plus and minus signs in the curl formula. So the left-hand side of the desired identity is
$$\varepsilon_{ijk}\partial_j (\varepsilon_{k\ell m}\partial_\ell A_m)$$
$$= \varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$
because derivatives commute with constants and with each other.
At this point, to simplify simplify $\varepsilon_{ijk}\varepsilon_{k\ell m}$, I just looked up the relevant identity of the Levi-Civita symbol, but it should be possible to derive it simply by algebraic manipulation from the purely combinatorial definition of the symbol. It turns out that $\varepsilon_{ijk}\varepsilon_{k\ell m} = \delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell}$, where $\delta_{ij}$ is the Kronecker delta which is $1$ if and only if $i = j$, and $0$ otherwise. This symbol acts like a substitution operator: $\delta_{ij}v_j = v_i$.
So we have
$$\varepsilon_{ijk}\varepsilon_{k\ell m}\partial_j\partial_\ell A_m$$
$$= (\delta_{i\ell}\delta_{jm} - \delta_{im}\delta_{j\ell})\partial_j\partial_\ell A_m$$
$$= \partial_m\partial_i A_m - \partial_\ell\partial_\ell A_i$$
$$= \partial_i (\partial_m A_m) - (\partial_\ell\partial_\ell) A_i.$$
The first term is $\nabla(\nabla\cdot A)$, while the second is $-(\nabla\cdot\nabla)A$.
