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...

Before proving this result,I prove a more general result:
Let $f(x,y,z)$ and $g(x,y,z)$ are two $3-$linear map on $\mathbb{R}^3$ with standard ordered basis $\{e_1,e_2,e_3\}$.
If $f(e_i,e_j,e_k)=g(e_i,e_j,e_k)$ for all $i,j,k \in \{1,2,3\}$,then $f(x,y,z)=g(x,y,z)$ for all $(x,y,z) \in (\mathbb{R}^3)^3$.
Proof:
For each $(x,y,z) \in (\mathbb{R}^3)^3$,let $x=\sum x_ie_i$, $y=\sum y_je_j$ and $z=\sum z_ke_k$.Then $f(x,y,z)=\sum_i \sum_j \sum_k x_iy_jz_kf(e_i,e_j,e_k)=\sum_i \sum_j \sum_k x_iy_jz_kg(e_i,e_j,e_k)=g(x,y,z)$.
Now,let $f(x,y,z)=(x×y)×z$,$g(x,y,z)=(x⋅z)y-(y⋅z)x$.It is not difficult to see that $f(e_i,e_j,e_k)=g(e_i,e_j,e_k)$.As a result, $f(x,y,z)=g(x,y,z)$ on $(\mathbb{R}^3)^3$ by the lemma.
Hence,$(x×y)×z=(x⋅z)y-(y⋅z)x$.
For your reference:Multilinear map
