Dot Product/ Cross Product Proof Let $\hat{a}$, $\hat{b}$, and $\hat{c}$  $\in \mathbb{R}^3$, using the properties of vectors, prove  $$ (\hat{a} \times \hat{b}) \cdot [(\hat{b} \times \hat{c}) \times (\hat{c} \times \hat{a})] = (\hat{a} \cdot (\hat{b} \times \hat{c}))^2$$
Proof:
Since $\hat{u} \cdot \hat{v} = \mid \mid  \hat{u}\mid \mid \mid \mid \hat{v} \mid \mid \cos \theta$ and $\mid \mid \hat{u} \times \hat{v} \mid \mid  = \mid \mid \hat{u} \mid \mid \mid \mid \hat{v} \mid \mid \sin \theta$, the above equation can be rewritten as
$$\mid \mid (\hat{a} \times \hat{b}) \mid \mid \mid \mid (\hat{b} \times \hat{c}) \times (\hat{c} \times \hat{a}) \mid \mid \cos \theta  = (\mid \mid \hat{a} \mid \mid \mid \mid (\hat{a} \times \hat{b}) \mid \mid \cos \theta)^2$$ which can be simplified further to $$\mid \mid \hat{a} \mid \mid\mid \mid \hat{b} \mid \mid \sin \theta \mid \mid \hat{b} \mid \mid \mid \mid \hat{c} \mid \mid \sin \theta \mid \mid \hat{c} \mid \mid \mid \mid \hat{a} \mid \mid \sin \theta \sin \theta \cos \theta  = (\mid \mid \hat{a} \mid \mid \mid \mid \hat{b} \mid \mid \mid \mid \hat{c} \mid \mid \sin \theta \cos \theta)^2 $$ reducing to 
$$ \mid \mid \hat{a} \mid \mid^2 \mid \mid \hat{b} \mid \mid^2 \mid \mid \hat{c} \mid \mid^2 \sin^2 \theta \sin^2 \theta \cos \theta =  \mid \mid \hat{a} \mid \mid^2 \mid \mid \hat{b} \mid \mid^2 \mid \mid \hat{c} \mid \mid^2 \sin^2 \theta \cos^2 \theta$$  
Please don't post an answer, I'm trying to work it on my own
let $\epsilon_{i_1i_2 \cdots i_n}$ be the permutation tensor for $n$ dimensions
$\hat{a} \times \hat{b} = \alpha_i e_i = \epsilon_{ijk} A_j B_k e_i$ ,
$\hat{b} \times \hat{c} = \beta_i e_i = \epsilon_{ijk} B_j C_k e_i$
$\hat{c} \times \hat{a} = \delta_i e_i = \epsilon_{ijk} C_j A_k e_i$  
then $(\hat{b} \times \hat{c}) \times (\hat{c} \times \hat{a}) = \gamma_i e_i = \epsilon_{ijk} \beta_j \delta_k e_i$ and
$$ (\hat{a} \times \hat{b}) \cdot [(\hat{b} \times \hat{c}) \times (\hat{c} \times \hat{a})] =\sum_{i=1}^{3}\alpha_i \gamma_i$$ so we get a new equation   $$\sum_{i=1}^{3}\alpha_i \gamma_i = (\hat{a} \cdot (\hat{b} \times \hat{c}))^2$$
$\hat{a} \cdot (\hat{b} \times \hat{c}) = \sum_{i=0}^{3} A_i \beta_i$ so we obtain  $$\sum_{i=1}^{3}\alpha_i \gamma_i  = \Big(\sum_{i=1}^{3} A_i \beta_i \Big)^2$$
How to proceed?
Just an fyi guys, this is just a problem in my Stewart Calculus book, so I don't think the proof should be that rigorous...
 A: Here is some hint to get you started:
$$\newcommand{\a}{\boldsymbol{a}}\newcommand{\b}{\boldsymbol{b}}\newcommand{\c}{\boldsymbol{c}}\newcommand{\u}{\boldsymbol{u}}\newcommand{\v}{\boldsymbol{v}}\newcommand{\w}{\boldsymbol{w}}
(\u\times \v)\times \w = (\u\cdot \w)\v - (\v\cdot \w)\u.\tag{$\star$}
$$
Now in the bracket you have:
$$
(\b\times \c)  \times  (\c\times \a) = \big(\b\cdot (\c\times\a)\big)\c - \big(\c\cdot (\c\times \a)\big)\b. 
$$
What is $\c\cdot (\c\times \a)$? Don't hover the mouse over the formula below if you wanna work by yourself first:

 $\c\cdot (\c\times \a) = - \a\cdot (\c\times \c) = 0$. Or you could argue that $\c\times \a$ produces a vector perpendicular to either $\a$ or $\c$.

Next step is to use (spoiler alert again) 

  $\b\cdot (\c\times\a) = \a\cdot (\b\times\c)$, then you have $$(\a \times \b)\cdot \big((\b\times \c)  \times  (\c\times \a) \big) = (\a \times \b)\cdot \Big(\big(\a\cdot (\b\times\c)\big)\c\Big). $$ Recalling $\a\cdot (\b\times\c)$ is a scalar, then what happened?

Formula $(\star)$ can be proved using index notation (easier), or a pure geometric proof like I said in the comments.
