1. **Problem:** Show that $$\nabla \times \left( \frac{\mathbf{a} \times \mathbf{r}}{|\mathbf{r}|^6} \right) = \frac{6 (\mathbf{a} \cdot \mathbf{r})}{|\mathbf{r}|^8} \mathbf{r} - \frac{4 \mathbf{a}}{|\mathbf{r}|^6}$$ where $\mathbf{a}$ is a constant vector and $\mathbf{r}$ is the position vector. 2. **Formula and rules:** - Use the vector calculus identity for curl of a product: $$\nabla \times (f \mathbf{F}) = f (\nabla \times \mathbf{F}) + (\nabla f) \times \mathbf{F}$$ - Here, $f = |\mathbf{r}|^{-6}$ and $\mathbf{F} = \mathbf{a} \times \mathbf{r}$. - Recall $\nabla \times (\mathbf{a} \times \mathbf{r}) = \mathbf{a} (\nabla \cdot \mathbf{r}) - (\mathbf{a} \cdot \nabla) \mathbf{r}$. - Since $\mathbf{a}$ is constant and $\nabla \cdot \mathbf{r} = 3$, and $(\mathbf{a} \cdot \nabla) \mathbf{r} = \mathbf{a}$. 3. **Intermediate work:** - Compute $\nabla \times \mathbf{F}$: $$\nabla \times (\mathbf{a} \times \mathbf{r}) = 3 \mathbf{a} - \mathbf{a} = 2 \mathbf{a}$$ - Compute $\nabla f$: $$\nabla |\mathbf{r}|^{-6} = -6 |\mathbf{r}|^{-8} \mathbf{r}$$ - Compute $ (\nabla f) \times \mathbf{F} $: $$(-6 |\mathbf{r}|^{-8} \mathbf{r}) \times (\mathbf{a} \times \mathbf{r}) = -6 |\mathbf{r}|^{-8} [\mathbf{r} \times (\mathbf{a} \times \mathbf{r})]$$ - Use vector triple product: $$\mathbf{r} \times (\mathbf{a} \times \mathbf{r}) = \mathbf{a} (\mathbf{r} \cdot \mathbf{r}) - \mathbf{r} (\mathbf{r} \cdot \mathbf{a}) = \mathbf{a} |\mathbf{r}|^2 - \mathbf{r} (\mathbf{a} \cdot \mathbf{r})$$ - Substitute back: $$-6 |\mathbf{r}|^{-8} (\mathbf{a} |\mathbf{r}|^2 - \mathbf{r} (\mathbf{a} \cdot \mathbf{r})) = -6 |\mathbf{r}|^{-6} \mathbf{a} + 6 |\mathbf{r}|^{-8} (\mathbf{a} \cdot \mathbf{r}) \mathbf{r}$$ 4. **Combine terms:** $$\nabla \times \left( \frac{\mathbf{a} \times \mathbf{r}}{|\mathbf{r}|^6} \right) = f (\nabla \times \mathbf{F}) + (\nabla f) \times \mathbf{F} = |\mathbf{r}|^{-6} (2 \mathbf{a}) + \left(-6 |\mathbf{r}|^{-6} \mathbf{a} + 6 |\mathbf{r}|^{-8} (\mathbf{a} \cdot \mathbf{r}) \mathbf{r} \right)$$ $$= 2 |\mathbf{r}|^{-6} \mathbf{a} - 6 |\mathbf{r}|^{-6} \mathbf{a} + 6 |\mathbf{r}|^{-8} (\mathbf{a} \cdot \mathbf{r}) \mathbf{r} = -4 |\mathbf{r}|^{-6} \mathbf{a} + 6 |\mathbf{r}|^{-8} (\mathbf{a} \cdot \mathbf{r}) \mathbf{r}$$ 5. **Final answer:** $$\boxed{\nabla \times \left( \frac{\mathbf{a} \times \mathbf{r}}{|\mathbf{r}|^6} \right) = \frac{6 (\mathbf{a} \cdot \mathbf{r})}{|\mathbf{r}|^8} \mathbf{r} - \frac{4 \mathbf{a}}{|\mathbf{r}|^6}}$$ This completes the proof.