Show that $$ \begin{aligned} \operatorname{div}[\mathbf{A}, \mathbf{B}] &=(\operatorname{curl} \mathbf{A}, \mathbf{B})-(\operatorname{curl} \mathbf{B}, \mathbf{A}), \\ \operatorname{curl} a \mathbf{A} &=\lfloor\operatorname{grad} a, \mathbf{A}]+a \text { curl } \mathbf{A}, \\ \operatorname{div} a \mathbf{A} &=(\operatorname{grad} a, \mathbf{A})+a \operatorname{div} \mathbf{A} . \end{aligned} $$

To prove the first property, we have: $$ \operatorname{div}[\mathbf{A}, \mathbf{B}] = (\operatorname{curl} \mathbf{A}, \mathbf{B}) - (\operatorname{curl} \mathbf{B}, \mathbf{A}) $$ We know that the divergence of the cross product of two vectors is given by: $$ \operatorname{div}[\mathbf{A}, \mathbf{B}] = \nabla \cdot (\mathbf{A} \times \mathbf{B}) $$ Now, let's expand the cross products in the right side of the property we are trying to prove: $$ (\operatorname{curl} \mathbf{A}, \mathbf{B}) = (\nabla \times \mathbf{A}) \cdot \mathbf{B} $$ and, $$ (\operatorname{curl} \mathbf{B}, \mathbf{A}) = (\nabla \times \mathbf{B}) \cdot \mathbf{A} $$ Using the curl of cross product identity \(\nabla \cdot (\mathbf{A} \times \mathbf{B}) = (\nabla \times \mathbf{A}) \cdot \mathbf{B} - (\nabla \times \mathbf{B}) \cdot \mathbf{A}\), we get: $$ \operatorname{div}[\mathbf{A}, \mathbf{B}] = (\operatorname{curl} \mathbf{A}, \mathbf{B}) - (\operatorname{curl} \mathbf{B}, \mathbf{A}) $$ Hence, the first property is proven.

To prove the second property, we need to show that: $$ \operatorname{curl} a \mathbf{A} = [\operatorname{grad} a, \mathbf{A}] + a \operatorname{curl} \mathbf{A} $$ We know that the curl operation is given by: $$ \operatorname{curl}(\mathbf{F}) = \nabla \times \mathbf{F} $$ Hence, $$ \operatorname{curl}(a\mathbf{A}) = \nabla \times (a\mathbf{A}) $$ Now, using the vector triple product identity \(\nabla \times (a\mathbf{A}) = (\nabla a) \times \mathbf{A} + a(\nabla \times \mathbf{A})\), we get: $$ \operatorname{curl}(a\mathbf{A}) = [\operatorname{grad} a, \mathbf{A}] + a \operatorname{curl} \mathbf{A} $$ Hence, the second property is proven.

To prove the third property, we need to show that: $$ \operatorname{div} a \mathbf{A} = (\operatorname{grad} a, \mathbf{A}) + a \operatorname{div} \mathbf{A} $$ We know that the divergence operation is given by: $$ \operatorname{div}(\mathbf{F}) = \nabla \cdot \mathbf{F} $$ Hence, $$ \operatorname{div}(a\mathbf{A}) = \nabla \cdot (a\mathbf{A}) $$ Now, using the product rule of dot products \(\nabla \cdot (a\mathbf{A}) = (\nabla a) \cdot \mathbf{A} + a(\nabla \cdot \mathbf{A})\), we get: $$ \operatorname{div}(a\mathbf{A}) = (\operatorname{grad} a, \mathbf{A}) + a \operatorname{div} \mathbf{A} $$ Hence, the third property is proven.