To prove the vector identity $(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D}) = (\mathbf{A} \cdot \mathbf{B} \times \mathbf{D}) \mathbf{C} - (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C}) \mathbf{D}$, we can use the Levi-Civita tensor $\varepsilon_{ijk}$ and Kronecker delta $\delta_{ij}$. Below is a step-by-step proof using these tools.

---

1. Expand each cross product in Levi-Civita notation

Cross product using the Levi-Civita symbol:

For two vectors $\mathbf{A} \times \mathbf{B}$, the $i$-th component is given by: $(\mathbf{A} \times \mathbf{B})_i = \varepsilon_{ijk} A_j B_k$ where the Levi-Civita symbol $\varepsilon_{ijk}$ is 1 or -1 depending on the permutation of indices, and 0 if any two indices are repeated.

Similarly, for the nested cross product $(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})$, the $i$-th component is: $[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \varepsilon_{ilm} (\mathbf{A} \times \mathbf{B})_l (\mathbf{C} \times \mathbf{D})_m$ Using the definition of cross products for each term: $(\mathbf{A} \times \mathbf{B})_l = \varepsilon_{lpq} A_p B_q, \quad (\mathbf{C} \times \mathbf{D})_m = \varepsilon_{mrs} C_r D_s$ Thus: $[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \varepsilon_{ilm} \varepsilon_{lpq} A_p B_q \varepsilon_{mrs} C_r D_s$

---

2. Use the Levi-Civita identity

We now use the Levi-Civita contraction identity: $\varepsilon_{ilm} \varepsilon_{lpq} = \delta_{ip} \delta_{mq} - \delta_{iq} \delta_{mp}$ Substitute this into the expression: $[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = (\delta_{ip} \delta_{mq} - \delta_{iq} \delta_{mp}) A_p B_q \varepsilon_{mrs} C_r D_s$ Distribute the terms: $[(\mathbf{A} \times \mathbf{B}) \times (\mathbf{C} \times \mathbf{D})]_i = \delta_{ip} \delta_{mq} A_p B_q \varepsilon_{mrs} C_r D_s - \delta_{iq} \delta_{mp} A_p B_q \varepsilon_{mrs} C_r D_s$

---

3. Simplify the expression

Using the properties of the Kronecker delta:

1. $\delta_{ip} A_p = A_i$
2. $\delta_{mq} B_q = B_m$
3. $\delta_{iq} B_q = B_i$
4. $\delta_{mp} A_p = A_m$

Thus, the first term becomes: $A_i B_m \varepsilon_{mrs} C_r D_s = A_i (\mathbf{B} \times \mathbf{D})_s C_r$