Vector calculus identity proof

Question

How can you prove this vector calculus identity?

$${\rm curl}(F\times G) = F {\rm div}(G) - G {\rm div}(F)+(G\cdot \nabla)F -(F\cdot \nabla)G $$

Answer 1

Abusing a bit Einstein's notation

\begin{eqnarray} \nabla \times (F \times G) &=& \hat{e}_i\epsilon_{ijk} \partial_j (F\times G)_k \\ &=& \hat{e}_i\epsilon_{ijk} \partial_j (\epsilon_{klm}F_l G_m) \\ &=& \hat{e}_i(\epsilon_{kij}\epsilon_{klm}) \partial_j (F_l G_m) \\ &=& \hat{e}_i (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})(\partial_jF_l G_m + F_l \partial_j G_m) \\ &=& \hat{e}_i \delta_{il}\delta_{jm}\partial_jF_lG_m + \hat{e}_i \delta_{il}\delta_{jm}F_l \partial_j G_m - \hat{e}_i \delta_{im}\delta_{jl}\partial_jF_l G_m - \hat{e}_i \delta_{im}\delta_{jl}F_l \partial_j G_m \\ &=& \hat{e}_i \partial_j F_i G_j + \hat{e}_i F_i \partial_j G_j - \hat{e}_i \partial_jF_jG_i - \hat{e}_i F_j\partial_jG_i \\ &=& (G_j\partial_j)(\hat{e}_iF_i) + (\hat{e}_i F_i)(\partial_j G_j) - (\hat{e}_iG_i)(\partial_j F_j) - (F_j \partial_j)(\hat{e}_iG_i) \\ &=& (G\cdot \nabla)(F) + F(\nabla\cdot G) - G(\nabla F) - (F\cdot\nabla)(G) \end{eqnarray}