How could index notation be used to prove this identity? How would you use index notation to prove that $\underline{\nabla} \times (\underline{u} \times \underline{v})=(\underline{\nabla} \cdot \underline{v}) \underline{u}-(\underline{\nabla} \cdot \underline{u}) \underline{v}+(\underline{v} \cdot \underline{\nabla}) \underline{u}-(\underline{u} \cdot \underline{\nabla}) \underline{v}$?
My attempt was as follows:
Starting with the left hand side of the equation, we have
\begin{align}
\underline{\nabla} \times (\underline{u} \times \underline{v})&=\varepsilon_{ijk}\frac{\partial}{\partial x_{j}}\varepsilon_{klm}u_{l}v_{m}=\varepsilon_{kij}\varepsilon_{klm}\frac{\partial}{\partial x_{j}}u_{l}v_{m}\\&
=(\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl})(v_{m} \frac{\partial u_{l}}{\partial x_{j}}+u_{l} \frac{\partial v_{m}}{\partial x_{j}}) \\ &=\delta_{il}\delta_{jm}v_{m} \frac{\partial u_{l}}{\partial x_{j}}+\delta_{il}\delta_{jm}u_{l} \frac{\partial v_{m}}{\partial x_{j}}-\delta_{im}\delta_{jl}v_{m} \frac{\partial u_{l}}{\partial x_{j}}-\delta_{im}\delta_{jl}u_{l} \frac{\partial v_{m}}{\partial x_{j}}\end{align}
However, I'm unsure of where to go from here. Any advice?
 A: Your approach is wrong from the first step itself since R.H.S. of 1st line is a scalar, but it should be a vector.
It should be as follows:
$$\nabla \times (u \times v) = \epsilon_{ijk} \frac{\partial}{\partial x_i}(u \times v)_j \hat e_k$$
$$=\epsilon_{ijk} \frac{\partial}{\partial x_i}(\epsilon_{lmj} \,\ u_l \,\ v_m ) \hat e_k$$
$$=\epsilon_{ijk} \epsilon_{lmj} \frac{\partial}{\partial x_i}( u_l \,\ v_m ) \hat e_k$$
$$=\epsilon_{kij} \epsilon_{lmj} \left[v_m\frac{\partial  u_l}{\partial x_i}+u_l\frac{\partial v_m}{\partial x_i}\right] \hat e_k$$
$$=(\delta_{kl} \delta_{im}-\delta_{km}\delta_{il}) \left[v_m\frac{\partial  u_l}{\partial x_i}+u_l\frac{\partial v_m}{\partial x_i}\right] \hat e_k$$
$$=\delta_{kl} \delta_{im}\left[v_m\frac{\partial  u_l}{\partial x_i}  \hat e_k+u_l\frac{\partial v_m}{\partial x_i}  \hat e_k\right]-\delta_{km}\delta_{il} \left[v_m\frac{\partial  u_l}{\partial x_i}  \hat e_k+u_l\frac{\partial v_m}{\partial x_i}  \hat e_k\right]$$
$$=\left[(v_m\frac{\partial }{\partial x_m})  (u_k \hat e_k) + (u_k \hat e_k)\frac{\partial v_i}{\partial x_i} \right]-\left[(v_k \hat e_k)\frac{\partial  u_i}{\partial x_i} + (u_i\frac{\partial }{\partial x_i}) (v_k \hat e_k)\right]$$
$$=(v \cdot \nabla)u+u(\nabla \cdot v)-v(\nabla \cdot u)-(u \cdot \nabla)v$$
$$=(\underline{\nabla} \cdot \underline{v}) \underline{u}-(\underline{\nabla} \cdot \underline{u}) \underline{v}+(\underline{v} \cdot \underline{\nabla}) \underline{u}-(\underline{u} \cdot \underline{\nabla}) \underline{v}$$
Hope this proof helps.
