Correct vector calculus divergence identity

I came across this vector calculus identity (in this wikipedia article, but also in other sources):

$$ \vec{\nabla} \cdot (\vec{B}\times \vec{C}) = \vec{C}\cdot(\vec{\nabla}\times \vec{B})-\vec{B}\cdot(\vec{\nabla}\times \vec{C}) $$ where $\vec{\nabla}=(\partial/\partial x,\partial/\partial y,\partial/\partial z)$. However, based on equation (2.5) of the Feynman Lectures on Physics, vol. II, which states that $$\vec{A} \cdot (\vec{B}\times \vec{C}) = (\vec{A}\times \vec{B})\cdot \vec{C},$$ the above identity should only contain the first term on the right-hand side.

I'm getting confused. Does anyone know which one is correct and what's the difference between the two?

The divergence operator looks like a vector but it's still an operator and contains derivatives. Most importantly you have to use the product rule, since the derivative acts on both $\mathbf{A}$ and $\mathbf{B}$. In general for these types of problems working with index notation and Einstein Summation is the best choice. For example \begin{align*} \nabla\cdot(\mathbf{A}\times\mathbf{B}) = \partial_i \varepsilon_{ijk} a_j b_k &= \varepsilon_{ijk} \left(\frac{\partial a_j}{\partial x_i} b_k + a_j \frac{\partial b_k}{\partial x_i}\right) \\ &= \varepsilon_{kij} b_k \partial_i a_j - \varepsilon_{jik} a_j \partial_i b_k \\ &= \mathbf{B}\cdot\left(\nabla\times\mathbf{A}\right) - \mathbf{A}\cdot\left(\nabla\times\mathbf{B}\right) \end{align*}