Answer
$div (F \times G)=G \cdot curl F-F \cdot curl G$
Work Step by Step
Need to prove that $div (F \times G)=G \cdot curl F-F \cdot curl G$
Let us consider that $F=A i+B j+C k$
Suppose $F=F_1i+F_2j+F_3k$ and $G=G_1i+G_2j+G_3k$
$div (F \times G)=\nabla \cdot (F \times G)$
$div (F \times G)=\dfrac{\partial}{\partial x}[G_3F_2-G_2F_3]-\dfrac{\partial}{\partial y}[G_3F_1-G_1F_3]+\dfrac{\partial}{\partial z}[G_2F_1-G_1F_2]$
$div (F \times G)=[G_1(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z})-G_2(\dfrac{\partial F_3}{\partial x}-\dfrac{\partial F_1}{\partial z})+G_3(\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y})]-[F_1(\dfrac{\partial G_3}{\partial y}-\dfrac{\partial G_2}{\partial z})-F_2(\dfrac{\partial G_3}{\partial x}-\dfrac{\partial G_1}{\partial z})+F_3(\dfrac{\partial G_2}{\partial x}-\dfrac{\partial G_1}{\partial y})]$
or, $div (F \times G)=(curl F) \cdot G-F \cdot (curl G)$
Hence, $div (F \times G)=G \cdot curl F-F \cdot curl G$
