Answer
See proof
Work Step by Step
Let: \[ \mathbf{F} = f(x, y, z)\mathbf{i} + g(x, y, z)\mathbf{j} + h(x, y, z)\mathbf{k}; \] \[ \mathbf{G} = P(x, y, z)\mathbf{i} + Q(x, y, z)\mathbf{j} + R(x, y, z)\mathbf{k} \] If \[ \mathbf{a} = a_{1}\mathbf{i} + a_{2}\mathbf{j} + a_{3}\mathbf{k} \text{ and } \mathbf{b} = b_{1}\mathbf{i} + b_{2}\mathbf{j} + b_{3}\mathbf{k} \] \[ \mathbf{a} \times \mathbf{b} = \left|\begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{matrix}\right| = \mathbf{i}(a_{2}b_{3} - a_{3}b_{2}) + \mathbf{j}(a_{3}b_{1} - a_{1}b_{3}) + \mathbf{k}(a_{1}b_{2} - b_{1}a_{2}) \] \[ \nabla \cdot \mathbf{a} = \left(\frac{da_{1}}{dx} + \frac{da_{2}}{dy} + \frac{da_{3}}{dz}\right) \] \[ \text{div}(\mathbf{F} + \mathbf{G}) = \left(\frac{\partial \mathbf{f}}{\partial \mathbf{x}} + \frac{\partial \mathbf{g}}{\partial \mathbf{y}} +\frac{\partial \mathbf{h}}{\partial \mathbf{z}}\right) +\left( \frac{\partial \mathbf{P}}{\partial \mathbf{x}}+ \frac{\partial \mathbf{Q}}{\partial \mathbf{y}} + \frac{\partial \mathbf{R}}{\partial \mathbf{z}}\right) = \text{div}(\mathbf{F}) + \text{div}(\mathbf{G}) \] Result: \[ \text{div}(\mathbf{F} + \mathbf{G}) = \text{div}(\mathbf{F}) + \text{div}(\mathbf{G}) \]
