1. **Problem:** Show that $\operatorname{div} \left( \frac{\mathbf{r}}{|\mathbf{r}|^3} \right) = 0$ where $\mathbf{r} = x \hat{i} + y \hat{j} + z \hat{k}$ and $r = |\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$. 2. **Formula and rules:** The divergence of a vector field $\mathbf{F} = F_x \hat{i} + F_y \hat{j} + F_z \hat{k}$ is given by $$\operatorname{div} \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}.$$ We will apply this to $\mathbf{F} = \frac{\mathbf{r}}{r^3} = \frac{x}{r^3} \hat{i} + \frac{y}{r^3} \hat{j} + \frac{z}{r^3} \hat{k}$. 3. **Calculate partial derivatives:** Calculate $\frac{\partial}{\partial x} \left( \frac{x}{r^3} \right)$: $$\frac{\partial}{\partial x} \left( \frac{x}{r^3} \right) = \frac{1}{r^3} - \frac{3x^2}{r^5}$$ Similarly, $$\frac{\partial}{\partial y} \left( \frac{y}{r^3} \right) = \frac{1}{r^3} - \frac{3y^2}{r^5}$$ $$\frac{\partial}{\partial z} \left( \frac{z}{r^3} \right) = \frac{1}{r^3} - \frac{3z^2}{r^5}$$ 4. **Sum the partial derivatives:** $$\operatorname{div} \left( \frac{\mathbf{r}}{r^3} \right) = \left( \frac{1}{r^3} - \frac{3x^2}{r^5} \right) + \left( \frac{1}{r^3} - \frac{3y^2}{r^5} \right) + \left( \frac{1}{r^3} - \frac{3z^2}{r^5} \right)$$ $$= \frac{3}{r^3} - \frac{3(x^2 + y^2 + z^2)}{r^5}$$ Since $x^2 + y^2 + z^2 = r^2$, $$= \frac{3}{r^3} - \frac{3r^2}{r^5} = \frac{3}{r^3} - \frac{3}{r^3} = 0.$$ 5. **Conclusion:** We have shown that $$\operatorname{div} \left( \frac{\mathbf{r}}{|\mathbf{r}|^3} \right) = 0.$$