1. The problem states the vector calculus identity: $$\nabla \times (\nabla \times \mathbf{F}) = \nabla (\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}$$ where $\mathbf{F}$ is a vector field. 2. This identity expresses the curl of the curl of $\mathbf{F}$ in terms of the gradient of the divergence of $\mathbf{F}$ minus the Laplacian of $\mathbf{F}$. 3. The curl of $\mathbf{F}$ is given by: $$\nabla \times \mathbf{F} = \left( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right)$$ 4. Applying the curl operator again to $\nabla \times \mathbf{F}$ yields $\nabla \times (\nabla \times \mathbf{F})$. 5. The gradient of the divergence of $\mathbf{F}$ is: $$\nabla (\nabla \cdot \mathbf{F}) = \left( \frac{\partial}{\partial x} (\nabla \cdot \mathbf{F}), \frac{\partial}{\partial y} (\nabla \cdot \mathbf{F}), \frac{\partial}{\partial z} (\nabla \cdot \mathbf{F}) \right)$$ 6. The Laplacian of $\mathbf{F}$ is the vector of Laplacians of each component: $$\nabla^2 \mathbf{F} = \left( \nabla^2 F_1, \nabla^2 F_2, \nabla^2 F_3 \right)$$ where $$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$ 7. The identity shows that the curl of the curl can be decomposed into these gradient and Laplacian terms, which is useful in vector calculus and physics, especially electromagnetism. Final answer: $$\boxed{\nabla \times (\nabla \times \mathbf{F}) = \nabla (\nabla \cdot \mathbf{F}) - \nabla^2 \mathbf{F}}$$