Answer
$div (F \times G)=G \cdot curl F-F \cdot curl G$
Work Step by Step
Plug $F=F_1i+F_2j+F_3k; G=G_1i+G_2j+G_3k$
$div (F \times G)=\nabla \cdot (F \times G)$
$div (F \times G)=\dfrac{\partial}{\partial x}[G_3F_2-G_2F_3]-\dfrac{\partial}{\partial y}[G_3F_1-G_1F_3]+\dfrac{\partial}{\partial z}[G_2F_1-G_1F_2]$
or, $=[(G_1)(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z})-(G_2)(\dfrac{\partial F_3}{\partial x}-\dfrac{\partial F_1}{\partial z})+(G_3) (\dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y})]-[(F_1) (\dfrac{\partial G_3}{\partial y}-\dfrac{\partial G_2}{\partial z})-(F_2) (\dfrac{\partial G_3}{\partial x}-\dfrac{\partial G_1}{\partial z})+(F_3) (\dfrac{\partial G_2}{\partial x}-\dfrac{\partial G_1}{\partial y})]$
or, $=(curl F) \cdot G-F \cdot (curl G)$
Hence, $div (F \times G)=G \cdot curl F-F \cdot curl G$ (Proved)
