For pure imaginary quaternions $u, v, w$, is there a way to prove the vector triple product $u\times(v\times w) = v(u\cdot w) - w(u\cdot v)$ from the relation: $$uv = -u\cdot v + u\times v \text{ for $u, v \in \mathbb{R}^3$ }$$
I tried many times but failed.
The vector triple product identity follows from quaternion associativity.
A convenient tool for expressing quaternion statements in terms of dot and cross products is to write the quaternion as a real number and a 3-vector, as $a = (\alpha, \mathbf{a})$. Then quaternion multiplication of quaternion $b = (\beta, \mathbf{b})$ on the left by $a$ is given by $$ a b = \begin{bmatrix} \alpha & -\mathbf{a}^T \\ \mathbf{a} & \alpha I + \mathbf{a} \times \end{bmatrix} \begin{pmatrix} \beta \\ \mathbf{b} \end{pmatrix} = \begin{pmatrix} \alpha\beta - \mathbf{a} \cdot \mathbf{b} \\ \beta\mathbf{a} + \alpha \mathbf{b} + \mathbf{a} \times \mathbf{b} \end{pmatrix} . $$
This is much simplified for purely non-real quaternions $a = (0,\mathbf{a})$, $b = (0,\mathbf{b})$ and $c = (0,\mathbf{c})$
$$ a b = \begin{bmatrix} 0 & -\mathbf{a}^T \\ \mathbf{a} & \mathbf{a} \times \end{bmatrix} \begin{pmatrix} 0 \\ \mathbf{b} \end{pmatrix} = \begin{pmatrix} -\mathbf{a} \cdot \mathbf{b} \\ \mathbf{a} \times \mathbf{b} \end{pmatrix} . $$ And further, $$ a b c = \begin{pmatrix} -(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \\ -(\mathbf{a} \cdot \mathbf{b}) \mathbf{c} + (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} \end{pmatrix} . $$
To extract the identity from quaternion expressions, for three cyclic permutations of three generic quaternions $a$, $b$ and $c$, write equations expressing associativity, to form a system of equations: $$ \left\{ \begin{array}{rcl} ( a b ) c - a ( b c ) &=& 0 \\ ( b c ) d - b ( c d ) &=& 0 \\ ( c a ) b - c ( a b ) &=& 0 . \end{array} \right. $$ In case the three quaternions are purely non-real, write $a = (0,\mathbf{a})$, $b = (0,\mathbf{b})$ and $c = (0,\mathbf{c})$, for which the above system implies $$ \left\{ \begin{array}{rcl} -( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c} +\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} ) + ( \mathbf{a} \times \mathbf{b} ) \times \mathbf{c} - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\ -( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a} +\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} ) + ( \mathbf{b} \times \mathbf{c} ) \times \mathbf{a} - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\ -( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b} +\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} ) + ( \mathbf{c} \times \mathbf{a} ) \times \mathbf{b} - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 . \end{array} \right. $$ The cross product terms can be compared more easily if the anticommutativity of the cross product is applied: $$ \left\{ \begin{array}{rcl} -( \mathbf{a} \cdot \mathbf{b} ) \mathbf{c} +\mathbf{a} ( \mathbf{b} \cdot \mathbf{c} ) - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) &=& 0 \\ -( \mathbf{b} \cdot \mathbf{c} ) \mathbf{a} +\mathbf{b} ( \mathbf{c} \cdot \mathbf{a} ) - \mathbf{a} \times ( \mathbf{b} \times \mathbf{c} ) - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) &=& 0 \\ -( \mathbf{c} \cdot \mathbf{a} ) \mathbf{b} +\mathbf{c} ( \mathbf{a} \cdot \mathbf{b} ) - \mathbf{b} \times ( \mathbf{c} \times \mathbf{a} ) - \mathbf{c} \times ( \mathbf{a} \times \mathbf{b} ) &=& 0 . \end{array} \right. $$ The identity results from adding the first two equations of the system, and subtracting the third.