Answer
See proof
Work Step by Step
Let $\vec{F} = x^3\mathbf{i} + y^3\mathbf{j} + z^3\mathbf{k}$ be a vector field. The divergence of this vector field is: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(z^3) = 3(x^2 + y^2 + z^2). \] Since $x^2 + y^2 + z^2 > 0$ for all $(x, y, z) \neq (0, 0, 0)$, then $\nabla \cdot \vec{F} > 0$ for all $(x, y, z) \neq (0, 0, 0)$. $\vec{F}$ has sources at all points except the origin where $\vec{F}$ doesn't have a source or sink.
