Does the denominator cancel? I recently asked this question: Simplify expression cross/dot products
The answer was straightforward, yet I can't apply the same to the expression below.
Can the following expression be simplified? In particular, I would like to remove the denominator:
$\frac{( b \times (a \times b)) \cdot (a \times (b\times a))}{(a\times b) \cdot (a\times b)}$
where $a$ and $b$ are 3 dimensional vectors.
Using indexes, the result seems to be:
$- a \cdot b$
but I fail to see how to simplify this using the properties of scalar and cross products.
 A: \begin{align}
\frac{(b\times(a\times b)) \cdot (a\times(b\times a))}{(a\times b) \cdot (a\times b)}
&= \frac{(b\times(a\times b)) \cdot (a\times(b\times a))}{\lVert a\times b\rVert^2}\\
&= \frac{1}{\lVert a\times b\rVert}(b\times(a\times b)) \cdot
   \frac{1}{\lVert a\times b\rVert}(a\times(b\times a))\\
&= \left(b\times\frac{1}{\lVert a\times b\rVert}(a\times b)\right) \cdot
   \left(a\times\frac{1}{\lVert a\times b\rVert}(b\times a))\right)\\
&= - (b\times c) \cdot (a\times c)
\end{align}
where $c = \frac{1}{\lVert a\times b\rVert}(a\times b)$
and the negative sign occurs when we replace 
$\frac{1}{\lVert a\times b\rVert}(b\times a)$ by 
$\frac{1}{\lVert a\times b\rVert}(a\times b).$
Note that $c$ is a unit vector perpendicular to both $a$ and $b,$ so 
$a\times c$ and $b\times c$ are just the vectors $a$ and $b$ rotated by
$90$ degrees (in the same direction) in the plane spanned by $a$ and $b.$
A: Your numerator is a dot product of two vector triple products. The former is equal to $(b \cdot b)a - (b \cdot a)b = \|b\|^2a - (a \cdot b)b$ and the latter is equal to $(a \cdot a)b - (a \cdot b)a = \|a\|^2b - (a \cdot b)a$, so their dot product is \begin{align*}(\|b\|^2a - (a\cdot b)b)\cdot(\|a\|^2b - (a\cdot b)a) &= \|a\|^2\|b\|^2(a \cdot b) + (a \cdot b)^3 - \|b\|^2(a\cdot a)(a \cdot b) - \|a\|^2(b \cdot b)(a \cdot b) \\&= (a \cdot b)^3 - \|a\|^2\|b\|^2(a \cdot b) \end{align*}
Your denominator is just $\|a \times b\|^2$, so
$$X := \dfrac{(b\times(a\times b))\cdot(a\times(b\times a))}{(a\times b)\cdot(a\times b)} = \dfrac{(a \cdot b)((a\cdot b)^2 - \|a\|^2\|b\|^2)}{\|a\times b\|^2}.$$
Now, if $\theta$ is the angle between $a$ and $b$, the numerator becomes $(a \cdot b)\|a\|^2\|b\|^2(\cos^2\theta - 1)$, while the denominator is $\|a\|^2\|b\|^2(\sin\theta)^2$, so the $\|a\|^2\|b\|^2$ cancel, giving $$X = \dfrac{(a\cdot b)(\cos^2\theta - 1)}{|\sin\theta|^2} = -a\cdot b.$$
