proof for $ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$ this formula just pop up in textbook I'm reading without any explanation
$ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$
I did some "vector arithmetic" using the determinant method
but I'm not getting the answer to agreed with above formula.
I'm wondering if anyone saw this formula before and know the proof of it?
the final result that i get is
$b_{1}(a_{3}c_{3}+a_{2}c_{2})-a_{1}(b_{2}c_{2}+b_{3}c_{3})i$
$b_{2}(a_{3}c_{3}+a_{1}c_{1})-a_{2}(b_{3}c_{3}+b_{1}c_{1})j$
$b_{3}(a_{2}c_{2}+a_{1}c_{1})-a_{3}(b_{2}c_{2}+b_{1}c_{1})k$
But I failed to see any correlation for $(\vec{A}\cdot\vec{C})$ and $(\vec{B}\cdot\vec{C})$ part...

$\def\A{{\bf A}}
\def\B{{\bf B}}
\def\C{{\bf C}}
\def\x{\times}
\def\o{\cdot}
\def\d{\delta}
\def\e{\varepsilon}$I have always found such products easiest to discover using the properties of the Kronecker delta and the Levi-Civita symbol. 
Note that $\A\o\B = A_i B_j \d_{ij}$ and $(\A\x\B)_k = A_i B_j \e_{ijk}$, 
where we use the Einstein summation convention. 
Also, $\e_{ijk}\e_{lmk} = \e_{ijk}\e_{klm} = \d_{il}\d_{jm} - \d_{im}\d_{jl}$. 
Then 
\begin{eqnarray*}
((\A\x\B)\x\C)_m &=& A_i B_j \e_{ijk} C_l \e_{klm} \\
&=& A_i B_j C_l(\d_{il}\d_{jm} - \d_{im}\d_{jl}) \\
&=& (\A\o\C)B_m - (\B\o\C)A_m.
\end{eqnarray*}
Therefore, 
$$(\A\x\B)\x\C = (\A\o\C)\B - (\B\o\C)\A,$$
as claimed. 
Addendum: 
In proving such identities, there is always the question of what formalism you have to work with.
My recommendation is to have the Levi-Civita symbol as it is more fundamental than the cross product. 
The result $\e_{ijk}\e_{lmk} = \d_{il}\d_{jm} - \d_{im}\d_{jl}$ is a consequence of the general properties of $\e$ but I give here a simple proof.
The product $\e_{ijk}\e_{lmk}$ is antisymmetric in $ij$ and $lm$. 
Thus, $\e_{ijk}\e_{lmk} = c \left(\d_{il}\d_{jm} - \d_{im}\d_{jl}\right)$.
But $\e_{123} = 1$, and so $1 = c\left(\d_{11}\d_{22} - \d_{12}\d_{21}\right)$. 
Therefore, $c=1$. 
