To solve the given vector identity, we aim to prove:

$\nabla \times (\nabla \times \mathbf{A}) = -\nabla^2 \mathbf{A} + \nabla (\nabla \cdot \mathbf{A}),$

where $\mathbf{A}$ is a vector field, $\nabla \times$ denotes the curl operator, $\nabla \cdot$ denotes the divergence operator, and $\nabla^2$ is the Laplacian operator.

Proof:

Step 1: Expand the double curl using vector calculus identity

The double curl of a vector field $\mathbf{A}$ can be expressed as:

$\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A}.$

This is a standard vector calculus identity.

Step 2: Verify terms in the equation

1. The term $\nabla (\nabla \cdot \mathbf{A})$ corresponds to the gradient of the divergence of $\mathbf{A}$.
2. The term $-\nabla^2 \mathbf{A}$ corresponds to the negative Laplacian applied to the vector field $\mathbf{A}$.

Hence, we have derived the required result:

$\nabla \times (\nabla \times \mathbf{A}) = -\nabla^2 \mathbf{A} + \nabla (\nabla \cdot \mathbf{A}).$

Conclusion:

The proof follows directly from the vector identity for the double curl. If you need more explanation or examples, let me know.

Would you like more details or step-by-step examples?

Here are some related questions to explore:

1. What is the physical significance of the Laplacian operator in vector fields?
2. How does the curl of a vector field relate to its rotational behavior?
3. Can we apply this identity to electromagnetic fields? If so, how?
4. What are the assumptions required for this vector identity to hold?
5. How can we generalize this result to non-Cartesian coordinates?

Tip: Memorizing standard vector identities is helpful in tackling problems involving electromagnetic theory, fluid dynamics, and advanced vector calculus!