Answer
See proof
Work Step by Step
Given the vector field $\vec{F} = (x^3 - x)\mathbf{i} + (y^3 - y)\mathbf{j} + (z^3 - z)\mathbf{k}$, whose divergence is given by: \[ \nabla \cdot \vec{F} = \frac{\partial}{\partial x}(x^3 - x) + \frac{\partial}{\partial y}(y^3 - y) + \frac{\partial}{\partial z}(z^3 - z) = 3(x^2 - 1) + 3(y^2 - 1) + 3(z^2 - 1) = 3(x^2 + y^2 + z^2 - 1), \] We note that $x^2 + y^2 + z^2 - 1 = 0$ is a sphere centered at the origin of radius $r = 1$. All points outside the sphere satisfy: \[ x^2 + y^2 + z^2 - 1 > 0 \Rightarrow \nabla \cdot \vec{F} > 0, \quad x^2 + y^2 + z^2 - 1 > 0 \Rightarrow \nabla \cdot \vec{F} > 0, \] In other words, all points outside the sphere are sources of the given vector field. Analogously, all points inside the sphere are sinks of the given vector field: \[ x^2 + y^2 + z^2 - 1 < 0 \Rightarrow \nabla \cdot \vec{F} < 0, \quad x^2 + y^2 + z^2 - 1 < 0 \Rightarrow \nabla \cdot \vec{F} < 0. \] Finally, on the surface of the sphere, $\vec{F}$ doesn't have sources or sinks.
