How can I prove/disprove $\vec A\times(\vec B\times\vec C)=(\vec A\times \vec B)\times\vec C +\vec B\times(\vec A\times\vec C)$ ?
I know I could equate the right side to: $\vec B(\vec A \cdot\vec C)-\vec C(\vec A \cdot\vec B)$ But I don't know where to go from there.
On
This is the Jacobi identity for the vector cross product. Since you already have the identity:
$$ \mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b}) $$
applying this to both sides should show you they are equivalent. The left-hand side is:
$$ LHS=\mathbf{a}\times(\mathbf{b}\times\mathbf{c})=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{a}\cdot\mathbf{b}) $$
and the right-hand side is:
$$ \begin{split} RHS&=(\mathbf{a}\times\mathbf{b})\times\mathbf{c}+\mathbf{b}\times(\mathbf{a}\times\mathbf{c}) \\ &=-\mathbf{c}\times(\mathbf{a}\times\mathbf{b})+\mathbf{a}(\mathbf{b}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a}) \\ &=-\mathbf{a}(\mathbf{c}\cdot\mathbf{b})+\mathbf{b}(\mathbf{c}\cdot\mathbf{a})+\mathbf{a}(\mathbf{b}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a}) \\ &=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{c}(\mathbf{b}\cdot\mathbf{a})=LHS \end{split} $$
The vector triple product satisfies the Jacobi identity: $$ a \times (b \times c) + b \times ( c \times a) + c \times (a \times b) =0, $$ because Lagrange's identity implies that the left-hand side expands to $$ b(c \cdot a)-c(a \cdot b) + c(a \cdot b)-a(b \cdot c)+a(b \cdot c)-b(c \cdot a), $$ and everything cancels. This is equivalent to your expression by the anticommutativity of the cross product, so it is always true.