Triple Cross Product Identity for imaginary quaternions $\mathbb{H}_0$

Question

Considering $x,y,z \in \mathbb{H}_0$, $x,y,z=\alpha$i $+ \beta $j $+ \gamma $k, prove the Triple Cross Product Identity: \begin{equation} (x \times y) \times z = y(x \bullet z) - x(y \bullet z) \end{equation} with the cross product defined as usual, and $(\alpha$i $+ \beta $j $+ \gamma $k$)\bullet(\alpha'$i $+ \beta' $j $+ \gamma' $k$)=\alpha\alpha'+\beta\beta'+\gamma\gamma'$.

The proof becomes straightforward if we define wlog $x=\alpha$i, $y=\alpha$i $+ \beta $j, $z=\alpha$i $+ \beta $j $+ \gamma $k, and I need to convert it to the desired form with Gram-Schmidt, but don't know exactly why or how. Any type of help is appreciated.

Answer 1

Using $$ x=x_1\,\mathbf{i}+x_2\,\mathbf{j}+x_3\,\mathbf{k}\,,\quad y=y_1\,\mathbf{i}+y_2\,\mathbf{j}+y_3\,\mathbf{k}\,, $$ it is straightforward to show that pure quaternions multiply as $$ xy=-x\bullet y+(x\times y)\,. $$ Since $x\bullet y$ is symmetric in $x,y$ and $x\times y$ is anti symmetric \begin{align} x\times y&=\frac{xy-yx}{2}\,,\quad x\bullet y=-\frac{xy+yx}{2}\,. \end{align} Then, clearly $$ (x\times y)\times z=\frac{xyz-yxz-zxy+zyx}{4}\,. $$ The RHS term $(x\bullet z)y-(y\bullet z)x$ is equal to this because it is

\begin{align}\require{cancel}&-\frac{xz+zx}{2}y+\frac{yz+zy}{2}x\\[2mm]&=-\frac{xz+zx}{4}y+\frac{yz+zy}{4}x-y\frac{xz+zx}{4}+x\frac{yz+zy}{4}\\[2mm]&=\frac{-\cancel{xzy}-zxy+\bcancel{yzx}+zyx-yxz+\bcancel{yzx}+xyz+\cancel{xzy}}{4}\,.\end{align}