I know you can use vector rules [BAC CAB and $ \vec a \cdot (\vec b \times \vec c) = \vec c \cdot (\vec a \times \vec b)$] on del operator as long as you keep track of what terms each $\nabla$ is acting on.
You can do this for $\nabla \times (\vec a \times \vec b) $:
$\nabla \times (\vec a \times \vec b) = \overset{\downarrow}{\vec a}(\nabla \cdot \overset{\downarrow}{\vec b}) - \overset{\downarrow}{\vec b} (\nabla \cdot \overset{\downarrow}{\vec a}) \\ = (\vec b \cdot \nabla) \overset{\downarrow}{\vec a} + \vec a (\nabla \cdot \overset{\downarrow}{\vec b}) - (\vec a \cdot \nabla) \overset{\downarrow}{\vec b} - \vec b(\nabla \cdot \overset{\downarrow}{\vec a}).$
For $\nabla \cdot (\vec a \times \vec b)$ you'll have:
$\nabla \cdot (\vec a \times \vec b) = \overset{\downarrow}{\vec b} \cdot (\nabla \times \overset{\downarrow}{\vec a})$
But how do say when $\nabla$ only acts on $\vec b$ the term will be $- \vec a \cdot (\nabla \times \overset{\downarrow}{\vec b})$?
Or for $\nabla(\vec a \cdot \vec b)$:
$\nabla(\vec a \cdot \vec b) = \overset{\downarrow}{\vec a} \times (\nabla \times \overset{\downarrow}{\vec b}) + \overset{\downarrow}{\vec b}(\overset{\downarrow}{\vec a} \cdot \nabla)$
How do proceed from here?
$ \newcommand\diffthis[1]{\overset{\downarrow}{#1}} $Let $L(v; a, b)$ be a function of vectors $v, a, b$ which is linear in $v$. Then as a generalization of the product rule $$ L(\nabla; \diffthis a, \diffthis b) = L(\nabla; \diffthis a, b) + L(\nabla; a, \diffthis b). $$ So for example $$ \nabla\cdot(\diffthis a\times\diffthis b) = \nabla\cdot(\diffthis a\times b) + \nabla\cdot(a\times\diffthis b). $$ Now using the triple product cycle identity the first term becomes $$ \nabla\cdot(\diffthis a\times b) = b\cdot(\nabla\times\diffthis a) $$ and the second term becomes $$ \nabla\cdot(a\times\diffthis b) = a\cdot(\diffthis b\times\nabla) = -a\cdot(\nabla\times\diffthis b) $$ so altogether $$ \nabla\cdot(\diffthis a\times\diffthis b) = b\cdot(\nabla\times\diffthis a) - a\cdot(\nabla\times\diffthis b) $$