Is there a way to prove the scalar triple product is invariant under cyclic permutations without using components? Proving that $$[a,b,c]=[b,c,a]=[c,a,b]$$ 
Where $$[a,b,c] = a \cdot (b \times c)$$
and $ a ,b, c$ are non-co-planar vectors, and is fairly easy when splitting them into component parts and doing it by brute force, however is there a more elegant way of doing this which doesn't involve this? I'm sure I've seen one in the past but I haven't been able to find it again. 
There is the geometric proof with the relation to parallelepipeds, although I feel this isn't a formal mathematical proof.
Any assistance would be appreciated!
 A: Expand the identity $$(a+b)\cdot((a+b)\times c) = 0$$ to obtain $$a\cdot(b\times c) = b \cdot(c\times a).$$
A: If you define $[a,b,c]$ as the determinant of a matrix with rows $a, b, c$, the property follows trivially.
A: The expressions $a\cdot (b\times c)$ and $b\cdot (c\times a)$ are linear in $a$, $b$ and $c$. Let $(e_1,e_2,e_3)$ denote the standard basis of ${\mathbb{R}}^3$. For $1\le n,m,r\le 3$, $e_n\cdot (e_m\times e_r) = e_m\cdot (e_r\times e_n)$. Hence the identity
$$a\cdot (b\times  c) = b\cdot (c\times a).$$
A cyclic permutation of the vectors $a,b$ and $c$ gives the other identity $b\cdot (c\times a) = c\cdot (a\times b).$
A: For $b\times c=0$ there's little to show.  Otherwise $\langle a,\frac{b\times c}{\|b\times c\|}\rangle$ is the component of $a$ orthogonal to the plane spanned by $b$ and $c$.  We further know that $\|b\times c\|$ is the area of the parallelogram spanned by $b$ and $c$.  From here we see that 
$$\langle a,\frac{b\times c}{\|b\times c\|}\rangle\|b\times c\|$$
is the signed volume of the parallelepiped spanned by $a$, $b$, and $c$, aka $\det(a,b,c)$.  Now apply the known rules of $\det$ to achieve the result.
A: Well, the required identity can be boiled down to the following observations. Let $V$ a 3 dimensional vector space over a field $F$, and write $\bigwedge^iV$ for its exterior powers.  
Make a choice of a basis vector $\omega$ for the 1 dimensional vector space $\bigwedge^3 V$, and of a non-degenerate bilinear form on $V$ (which we'll write as a "dot product").
The choice of the basis vector $\omega$ amounts to identifying $\bigwedge^2 V$ with the dual $V^\vee$. 
The choice of the non-degenerate bilinear form on $V$ thus identifies $V$ and $V^\vee = \bigwedge^2 V$.
So: given $\mathbf{a_j} \in V$  for $j=1,2,3$, we view
$\mathbf{a_2} \times \mathbf{a_3}$ as the vector in $V$ identified as above
with the vector $\mathbf{a_2} \wedge \mathbf{a_3} \in \bigwedge^2 V$.
With this understanding, the "scalar triple product" $\mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a_3}) \in F$ for $\mathbf{a_j} \in V$ may be described by
$$\mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a_3})\omega = \mathbf{a_1} \wedge \mathbf{a_2} \wedge \mathbf{a_3} \in {\bigwedge}^3 V = F\omega$$
And so for any even permutation $\sigma$ of $1,2,3$ -- in particular, a cyclic permutation -- the skew commutativity of the exterior algebra $\bigwedge V$ implies that 
$$\mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a_3})\omega =
\mathbf{a_1} \wedge \mathbf{a_2} \wedge \mathbf{a_3} =
\mathbf{a_{\sigma(1)}} \wedge \mathbf{a_{\sigma(2)}} \wedge \mathbf{a_{\sigma(3)}} =
\mathbf{a_{\sigma(1)}} \cdot (\mathbf{a_{\sigma(2)}} \times \mathbf{a_{\sigma(3)}})\omega$$
so that
$$\mathbf{a_1} \cdot (\mathbf{a_2} \times \mathbf{a_3})=
\mathbf{a_{\sigma(1)}} \cdot (\mathbf{a_{\sigma(2)}} \times \mathbf{a_{\sigma(3)}}).$$
