1. **State the problem:** We want to show that $$\nabla \times \left( \frac{\vec{a} \times \vec{x}}{r^3} \right) = -\frac{\vec{a}}{r^3} + 3 \frac{(\vec{a} \cdot \vec{x}) \vec{x}}{r^5}$$ where $\vec{a}$ is a constant vector, $\vec{x} = (x,y,z)$ is the position vector, and $r = |\vec{x}| = \sqrt{x^2 + y^2 + z^2}$. 2. **Recall vector calculus identities:** We will use the identity for the curl of a cross product: $$\nabla \times (\vec{A} \times \vec{B}) = \vec{A}(\nabla \cdot \vec{B}) - \vec{B}(\nabla \cdot \vec{A}) + (\vec{B} \cdot \nabla)\vec{A} - (\vec{A} \cdot \nabla)\vec{B}$$ 3. **Set vectors:** Let $$\vec{A} = \vec{a}$$ (constant vector, so $\nabla \cdot \vec{A} = 0$, and $(\vec{B} \cdot \nabla)\vec{A} = 0$) Let $$\vec{B} = \frac{\vec{x}}{r^3}$$ 4. **Calculate terms:** - $\nabla \cdot \vec{B} = \nabla \cdot \left( \frac{\vec{x}}{r^3} \right)$ Since $\vec{x} = (x,y,z)$, and $r^2 = x^2 + y^2 + z^2$, Using divergence formula: $$\nabla \cdot \left( \frac{\vec{x}}{r^3} \right) = \frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) + \frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) + \frac{\partial}{\partial z} \left( \frac{z}{r^3} \right)$$ By symmetry each term is similar; compute one: $$\frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) = \frac{1}{r^3} - 3 \frac{x^2}{r^5}$$ Summing all three variables: $$\nabla \cdot \vec{B} = 3 \frac{1}{r^{3}} - 3 \frac{x^{2} + y^{2} + z^{2}}{r^{5}} = 3 \frac{1}{r^{3}} - 3 \frac{r^{2}}{r^{5}} = 3 \frac{1}{r^{3}} - 3 \frac{1}{r^{3}} = 0$$ So, $\nabla \cdot \vec{B} = 0$. 5. **Since $\vec{a}$ is constant, $\nabla \cdot \vec{A}=0$ and $(\vec{B} \cdot \nabla)\vec{A}=0$.** 6. **Calculate $(\vec{A} \cdot \nabla)\vec{B}$:** Since $\vec{a}$ is constant, $$(\vec{a} \cdot \nabla) \vec{B} = a_i \frac{\partial}{\partial x_i} \left( \frac{x_j}{r^{3}} \right) \hat{e}_j$$ Calculate $\frac{\partial}{\partial x_i} \left( \frac{x_j}{r^{3}} \right)$: $$\frac{\partial}{\partial x_i} \left( x_j r^{-3} \right) = \delta_{ij} r^{-3} + x_j \frac{\partial}{\partial x_i} r^{-3}$$ But $\frac{\partial}{\partial x_i} r^{-3} = -3 r^{-5} x_i$, So: $$\frac{\partial}{\partial x_i} \left( \frac{x_j}{r^{3}} \right) = \delta_{ij} r^{-3} - 3 x_j r^{-5} x_i$$ Thus, $$(\vec{a} \cdot \nabla) \vec{B} = a_i ( \delta_{ij} r^{-3} - 3 x_j r^{-5} x_i ) \hat{e}_j = r^{-3} a_j \hat{e}_j - 3 r^{-5} x_j x_i a_i \hat{e}_j = \frac{\vec{a}}{r^{3}} - 3 \frac{(\vec{a} \cdot \vec{x}) \vec{x}}{r^{5}}$$ 7. **Combine terms:** Using the curl identity: $$\nabla \times \left( \vec{a} \times \frac{\vec{x}}{r^3} \right) = \vec{a} (\nabla \cdot \vec{B}) - \vec{B} (\nabla \cdot \vec{a}) + (\vec{B} \cdot \nabla) \vec{a} - (\vec{a} \cdot \nabla) \vec{B}$$ Substituting the terms: $$= \vec{a} (0) - \frac{\vec{x}}{r^3} (0) + 0 - \left( \frac{\vec{a}}{r^{3}} - 3 \frac{(\vec{a} \cdot \vec{x}) \vec{x}}{r^{5}} \right) = -\frac{\vec{a}}{r^{3}} + 3 \frac{(\vec{a} \cdot \vec{x}) \vec{x}}{r^{5}}$$ 8. **Conclusion:** We have shown the identity as required: $$\nabla \times \left( \frac{\vec{a} \times \vec{x}}{r^{3}} \right) = -\frac{\vec{a}}{r^{3}} + 3 \frac{(\vec{a} \cdot \vec{x}) \vec{x}}{r^{5}}$$