Triple quaternion multiplication I'm self learner and for some reason I can't wrap my head around quaternion multiplication. 
I just stumble upon one of equation in my text.
Can anyone show step-by-step workout for below:
$$
\begin{equation}
\begin{split}
w =& qvq^* \\
  =& (q_0+\vec{q})(0+\vec{v})(q_0-\vec{q}) \\  
  =&(2q_0^2-1)\vec{v}+2(\vec{q}\cdot\vec{v})\vec{q} + 2q_0(\vec{q}\times\vec{v})
\end{split}
\end{equation}
$$
 A: Here's one way to do it. 
$\def\vq{{\bf q}}
\def\vqb{{\bf q}^*}
\def\vv{{\bf v}}
\def\va{{\bf a}}
\def\vb{{\bf b}}
\def\vc{{\bf c}}$
$$\begin{eqnarray*}
q v q^* &=& (v q + [q,v])q^* \\
&=& v \|q\|^2
    + 2(\vq\times\vv) (q_0 - \vq) \\
&=& v \|q\|^2
    + 2q_0(\vq\times\vv) - 2(\vq\times\vv)\vq \\
&=& v \|q\|^2
    + 2q_0(\vq\times\vv) 
    - 2\left[-(\vq\times\vv)\cdot\vq + (\vq\times\vv)\times\vq \right]\\
&=& v \|q\|^2
    + 2q_0(\vq\times\vv) 
    + 2\vq\times(\vq\times\vv) \\
&=& v \|q\|^2
    + 2q_0(\vq\times\vv) 
    + 2\vq(\vq\cdot\vv) 
    - 2\vv(\vq\cdot\vq) \\
&=& v_0\|q\|^2 + (q_0^2-\vq\cdot\vq)\vv 
    + 2(\vq\cdot\vv)\vq 
    + 2q_0(\vq\times\vv) \\
&=& (2q_0^2-1)\vv 
    + 2(\vq\cdot\vv)\vq 
    + 2q_0(\vq\times\vv), \hspace{5ex} \textrm{if }v_0=0\textrm{ and }\|q\| = 1
\end{eqnarray*}$$
Some relevant links and formulas: 
$$\begin{eqnarray*}
a b &=& (a_0+\va)(b_0+\vb) \\
    &=& a_0b_0-\va\cdot\vb + a_0\vb+b_0\va+\va\times\vb \\ 
\hspace{0ex}[a,b] &=& ab-ba \\
    &=& 2\va\times\vb \\
\|a\|^2 &=& a_0^2 + \va\cdot\va \\
    &=& 1, \textrm{ for unit quaternion} \\
\va\cdot(\vb\times\vc) &=& \vb\cdot(\vc\times\va) \\
    &=& \vc\cdot(\va\times\vb) \\
\va\times\va &=& 0 \\
\va\times(\vb\times\vc) &=& \vb(\va\cdot\vc) - \vc(\va\cdot\vb) 
\end{eqnarray*}$$
