Curl of vector product with constant and position vector in index notation

Question

I need to show that for a given constant vector C and position vector R

Curl [ (C × R) ×R ] = 3C × R

I treat R as x_ie_i. This is what I've tried so far but the second position vector R is throwing me off:

\begin{align*} \nabla \times ((\mathbf{C} \times \mathbf{R}) \times \mathbf{R}) &= \epsilon_{ijk} \partial_i ((\mathbf{C} \times \mathbf{R}) \times \mathbf{R})_j \mathbf{e}_k \\ &= \epsilon_{ijk} \partial_i (\epsilon_{\ell mj} (\mathbf{C} \times \mathbf{R})_\ell x_m)\mathbf{e}_k \\ &= \epsilon_{ijk} \partial_i (\epsilon_{\ell mj} \epsilon_{rp\ell}x_r c_p x_m) \mathbf{e}_k\\ &= \epsilon_{ijk} \epsilon_{\ell mj} \epsilon_{rp\ell} \partial_i x_r x_m c_p \mathbf{e}_k\\ &= \epsilon_{ijk} \epsilon_{\ell mj} \epsilon_{rp\ell} \delta_{ir} c_p x_m \mathbf{e}_k\\ &= \epsilon_{ijk} \epsilon_{\ell mj} \epsilon_{ip\ell} c_p x_m \mathbf{e}_k\\ \end{align*} I think I'm going wrong on the delta line where I make kronecker delta ir, but the x_m is still there. After this I tried using the contraction identity on the first and last symbol because they both start with i $$ \epsilon_{ijk}\epsilon_{ip\ell} = \delta_{jp}\delta_{k\ell} - \delta_{j\ell}\delta_{kp} $$ This leads onto: \begin{align*} \epsilon_{\ell mj}\delta_{jp}\delta_{k\ell} &= \epsilon_{kmp} \\ \epsilon_{\ell mj}\delta_{j\ell}\delta_{kp} &= \epsilon_{\ell m\ell} = 0 \end{align*}

Which doesn't help me solve the equation so I'm trying to figure out where I went wrong

Answer 1

Using direct definition, anyway it works :

Let $C=(c_1,c_2,c_3)$ and $R=(x_1,x_2,x_3)$. Then $C\times R=(c_2x_3-c_3x_2,c_3x_1-c_1x_3,c_1x_2-c_2x_1)$ $$(C\times R)\times R=(*,(c_1x_2-c_2x_1)x_3-(c_2x_3-c_3x_2)x_3,(c_2x_3-c_3x_2)x_2-(c_3x_1-c_1x_3)x_1)$$

Then $[\nabla((C\times R)\times R)]_1=-c_3x_2+(c_2x_3-c_3x_2)-[-c_2x_3-(c_2x_3-c_3x_2)]=3(c_2x_3-c_3x_2)=3[C\times R]_1$

Hope this helps a bit.