Show that $\nabla \times (\mathbf{c}\times \mathbf{r}) = 2\mathbf{c}$

Question

Suppose $r=x_i\mathbf{e}_i$

I need to do this question using the Einstein summation convention. So far I have:

$\nabla \times (\mathbf{c}\times \mathbf{r}) = \varepsilon_{ijk} \frac{\partial}{\partial x_j } \varepsilon_{kmn}c_mx_n$

$=\varepsilon_{ijk}\varepsilon_{kmn}c_m \frac{\partial x_n}{\partial x_j}$

$= \varepsilon_{ijk}\varepsilon_{kmn} c_m \delta_{nj}$

$=\varepsilon_{jki}\varepsilon_{jkm}c_m$

$=(\delta_{kk} \delta_{im} - \delta_{km}\delta_{ik})c_m$

$=\delta_{kk}\delta_{im}c_m - \delta_{km}\delta_{ik}c_m$

$=\delta_{kk}c_i - \delta_{ik}c_m$

Now I'm stuck. Thanks for help in advance!

Answer 1

All correct up to your last line. The second term should be $\delta_{im}c_m=c_i$
and remember that $\delta_{kk}=3$ in three-dimensional space.