Answer
$$\operatorname{div}(\mathbf{F} \times \mathbf{G}) = \mathbf{G} \cdot \operatorname{curl}(\mathbf{F})-\mathbf{F} \cdot \operatorname{curl}(\mathbf{G})$$
Work Step by Step
We verify as follows:
\begin{align*}
\operatorname{div}(\mathbf{F} \times \mathbf{G})&= \begin{array}{|||}
\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
\mathbf{F}_1& \mathbf{F}_2&\mathbf{F}_3\\
\mathbf{G}_1& \mathbf{G}_2&\mathbf{G}_3
\end{array}
\\
&=\frac{\partial}{\partial x}\left( \mathbf{F}_{2} \mathbf{G}_{3}- \mathbf{F}_{3} \mathbf{G}_{2}\right)+\frac{\partial}{\partial y}\left( \mathbf{F}_{3}\mathbf{G}_{1}- \mathbf{F}_{1} \mathbf{G}_{3}\right)+\frac{\partial}{\partial z}\left( \mathbf{F}_{1} \mathbf{G}_{2}- \mathbf{F}_{2} \mathbf{G}_{1}\right)\\
&= \mathbf{F}_{1}\left(\frac{\partial \mathbf{G}_{2}}{\partial z}-\frac{\partial \mathbf{G}_{3}}{\partial y}\right)+\mathbf{F}_{2}\left(\frac{\partial \mathbf{G}_{3}}{\partial x}-\frac{\partial \mathbf{G}_{1}}{\partial z}\right)
+\mathbf{F}_{3}\left(\frac{\partial \mathbf{G}_{1}}{\partial y}-\frac{\partial \mathbf{G}_{2}}{\partial z}\right)+ \mathbf{G}_{1}\left(\frac{\partial \mathbf{F}_{3}}{\partial y}-\frac{\partial \mathbf{F}_{2}}{\partial z}\right)
+ \mathbf{G}_{2}\left(\frac{\partial \mathbf{F}_{1}}{\partial z}-\frac{\partial \mathbf{F}_{3}}{\partial x}\right)+ \mathbf{G}_{3}\left(\frac{\partial \mathbf{F}_{2}}{\partial x}-\frac{\partial \mathbf{F}_{1}}{\partial y}\right) \\
&= \mathbf{G} \cdot \operatorname{curl}(\mathbf{F})-\mathbf{F} \cdot \operatorname{curl}(\mathbf{G})
\end{align*}
