Is there a way to prove vector triple product from quaternion multiplication? 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.
 A: 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.
