While trying to prove
$$[c\cdot (b\cdot\nabla) - b\cdot(c\cdot\nabla)]a = (\nabla\times a) \cdot (b\times c)$$
I had some difficulties on how to treat the term $(b\cdot\nabla)$. It seems that $(b\cdot\nabla)$ is an operator. But how to use it? And also any one can help in prooving the above formula?
Let $e_1,e_2,e_3$ be your (orthonormal) basis vectors so that any vector $x = (x_1,x_2,x_3)$ can be written $x = x_1e_1 + x_2e_2 + x_3e_3$. The basis vectors, usually $e_1=(1,0,0)$, $e_2=(0,1,0)$ and $e_3=(0,0,1)$ satisfy $e_i \cdot e_j = 0$ if $i\not=j$ and $e_i\cdot e_i = 1$. Using this we have that the operator $b\cdot\nabla$ can be written
$$b\cdot \nabla = (b_1e_1+b_2e_2+b_3e_3)\cdot\left(\frac{\partial}{\partial x_1} e_1+\frac{\partial}{\partial x_1} e_2+\frac{\partial}{\partial x_1} e_3\right) \\= b_1\frac{\partial}{\partial x_1} + b_2\frac{\partial}{\partial x_2} + b_3\frac{\partial}{\partial x_3}$$
which is a scalar operator (in other words it is not a vector itself). When this operator acts on a vector $a= a_1 e_1 + a_2 e_2 + a_3e_3$ then the result is a vector
$$[b\cdot \nabla]a = \left(b_1\frac{\partial a_1}{\partial x_1} + b_2\frac{\partial a_1}{\partial x_2} + b_3\frac{\partial a_1}{\partial x_3}\right)e_1 \\ + \left(b_1\frac{\partial a_2}{\partial x_1} + b_2\frac{\partial a_2}{\partial x_2} + b_3\frac{\partial a_2}{\partial x_3}\right)e_2+\left(b_1\frac{\partial a_3}{\partial x_1} + b_2\frac{\partial a_3}{\partial x_2} + b_3\frac{\partial a_3}{\partial x_3}\right)e_3$$
In more compact form the $i$'th component can be written
$$([b\cdot \nabla]a)_i = b_1\frac{\partial a_i}{\partial x_1} + b_2\frac{\partial a_i}{\partial x_2} + b_3\frac{\partial a_i}{\partial x_3}$$
or even more compact by using a sum $([b\cdot \nabla]a)_i = \sum_{j=1}^3b_j \frac{\partial a_i}{\partial x_j}$.
To prove the formula you can simply write out both the left and right hand side of the equation in terms of the components of $a,b,c$ and compare the two expressions (which should be equal for any value of the components of $a,b,c$).