Proof of identity: cross product of three vectors A book I'm reading contains the following (paraphrased)
\begin{equation}
(a \times b) \times c = (a \cdot c)b - (b \cdot c)a
\end{equation}
This is supposed to follow from:
\begin{equation}
(a \times b) \cdot (c \times d) = (a \cdot c)(b \cdot d) - (a \cdot d)(b \cdot c)
\end{equation}
Where in both equations $a, b, c, d$ are all vectors. The question is how to prove this? 
 A: If you know the identity
$$x \cdot (y \times z) = (x \times y) \cdot z$$
for arbitrary vectors $x,y,z$, then putting $x=a \times b$ and $y=c$, $z=d$, you have
$$ (a \times b) \cdot (c \times d) = ((a \times b) \times c) \cdot d $$
so you can arrange your second equation so that everything is dotted with $d$:
$$ ((a \times b) \times c) \cdot d = ((a \cdot c)b-(b \cdot c)a) \cdot d $$
and $d$ is arbitrary, so the vectors dotted on both sides must be the same.
A: Chappers' proof is the way that was most likely intended (using the identity given). However, here is another approach (if only to show how much work is saved by using the identity given)
Since  $(a\times b)\times a$ is perpendicular to $a$,
$$
((a\times b)\times a)\cdot a=0\tag{1}
$$
Since $|a\times b|=|a|\,|b|\,|\sin(\theta)|$ and $a\cdot b=|a|\,|b|\cos(\theta)$, we have
$$
|a\times b|^2+|a\cdot b|^2=|a|^2|b|^2\tag{2}
$$
Since $(a\times b)\cdot c=(b\times c)\cdot a$, we can use $(2)$ to get
$$
\begin{align}
(\color{#C00000}{(a\times b)}\times\color{#00A000}{a})\cdot\color{#0000F0}{b}
&=(\color{#00A000}{a}\times \color{#0000F0}{b})\cdot\color{#C00000}{(a\times b)}\\
&=(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)\tag{3}
\end{align}
$$
$(a\times b)\times a$ is perpendicular to $a\times b$ which is perpendicular to both $a$ and $b$. Therefore, $(a\times b)\times a$ is in the plane of $a$ and $b$. Thus, we can verify the following equation by taking dot products with $a$ and $b$ and using $(1)$ and $(3)$
$$
(a\times b)\times a=(a\cdot a)b-(a\cdot b)a\tag{4}
$$
Swapping $a$ and $b$ and negating $(4)$ gives
$$
(a\times b)\times b=(a\cdot b)b-(b\cdot b)a\tag{5}
$$
Suppose that $c$ is in the plane of $a$ and $b$. Writing $c=xa+yb$, we can take the dot product of this equation with $a$ and $b$ to get two equations to solve for $x$ and $y$. This gives
$$
c=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}a+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}b\tag{6}
$$
Furthermore, since $c$ only appears on the right side of $(6)$ in a dot product with $a$ or $b$, any component of $c$ perpendicular to the plane of $a$ and $b$ will not change the right side. Thus, the right side of $(6)$ is a formula for the perpendicular projection of $c$ onto the plane of $a$ and $b$.
Using $(4)$, $(5)$, and $(6)$, we have
$$
\begin{align}
(a\times b)\times c
&=\tfrac{(a\cdot c)(b\cdot b)-(b\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot a)b-(a\cdot b)a)\\
&+\tfrac{(a\cdot a)(b\cdot c)-(a\cdot c)(a\cdot b)}{(a\cdot a)(b\cdot b)-(a\cdot b)(a\cdot b)}((a\cdot b)b-(b\cdot b)a)\\[4pt]
&=(a\cdot c)b-(b\cdot c)a\tag{7}
\end{align}
$$
