proof for $ (\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A}\cdot\vec{C})\vec{B}-(\vec{B}\cdot\vec{C})\vec{A}$

Let $a=\begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix}$ and likewise for $b,c.$ The first component of $a\times (b \times c)$ is seen by applying the definition to be $a_2(b_1c_2 - b_2c_1) - a_3(b_3c_1-b_1c_3).$ With some algebra, this is seen to be $(a_2c_2+a_3c_3)b_1 - (a_2b_2+a_3b_3)c_1.$ The tricky transformation to apply here is simulataneously adding and subtracting the quantity $a_1b_1c_1.$ This allows us now to write the first component as $(a_1c_1 + a_2c_2 +a_3c_3)b_1 - (a_1b_1 + a_2b_2 + a_3b_3)c_1.$ Applying similar arguments to the second and third coordinates, we see that $a\times (b\times c) = \begin{pmatrix}(a_1c_1 + a_2c_2 +a_3c_3)b_1 - (a_1b_1 + a_2b_2 + a_3b_3)c_1 \\ (a_1c_1 + a_2c_2 +a_3c_3)b_2 - (a_1b_1 + a_2b_2 + a_3b_3)c_2 \\ (a_1c_1 + a_2c_2 +a_3c_3)b_3 - (a_1b_1 + a_2b_2 + a_3b_3)c_3 \end{pmatrix} = (a\cdot c)b - (a\cdot b)c$

vector $ \vec A\times \vec B$ is perpendicular to the plane containing $\vec A $ and $\vec B$.Now, $(\vec A\times \vec B)\times \vec C$ is perpendicular to plane containing vectors $\vec C$ and $\vec A\times \vec B$, thus $(\vec A\times \vec B)\times \vec C$ is in the plane containing $\vec A$ and $ \vec B$ and hence $(\vec A\times \vec B)\times \vec C$ is a linear combination of vectors $\vec A$ and $ \vec B\implies (\vec A\times \vec B)\times \vec C=\alpha \vec A + \beta \vec B$. Take dot product with $\vec C$ both sides gives $0$ on the L.H.S(as $(\vec A\times \vec B)\times \vec C$ is perpendicular to $\vec C$), hence $0=\alpha (\vec A. \vec C)+\beta(\vec B.\vec C)\implies \frac{\beta}{\vec A. \vec C}=\frac{-\alpha}{\vec B. \vec C}=\lambda \implies \alpha=-\lambda(\vec B. \vec C)$ and $\beta=\lambda(\vec A. \vec C) \implies (\vec A\times \vec B)\times \vec C=\lambda((\vec A. \vec C)\vec B-(\vec B. \vec C)\vec A)$. Here, $\lambda$ is independent of the magnitudes of vectors as if magnitude of any vector is multiplied by any scalar, that appears explicitly on both sides of the equation.Thus , putting unit vectors $\vec i,\vec j,\vec i$ in the equation gives $\vec j=\lambda(\vec j)\implies \lambda=1$ and hence $(\vec A\times \vec B)\times \vec C=((\vec A. \vec C)\vec B-(\vec B. \vec C)\vec A)$

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