Multiplication of determinants Show that for any vectors $\bf{a}$,$\bf{b}$,$\bf{c}$,$\bf{u}$,$\bf{v}$,$\bf{w}$ in $\mathbb{R}^3$,
$$[\bf{a},\bf{b},\bf{c}][\bf{u},\bf{v},\bf{w}]=\begin{vmatrix}\bf{a}\cdot\bf{u}&\bf{a}\cdot\bf{v}&\bf{a}\cdot\bf{w}\\
\bf{b}\cdot\bf{u}&\bf{b}\cdot\bf{v}&\bf{b}\cdot\bf{w}\\
\bf{c}\cdot\bf{u}&\bf{c}\cdot\bf{v}&\bf{c}\cdot\bf{w}\\
\end{vmatrix}$$
My question: is there a more elegant way to prove that without actually delving into tedious direct approach by definition?
I was considering that "summing" approach, using the fact that
$$[\textbf{a},\textbf{b},\textbf{c}][\textbf{u},\textbf{v},\textbf{w}]=\bigg(\sum^3_{i,j,k=1}\epsilon_{ijk}a_ib_jc_k\bigg)\bigg(\sum^3_{i,j,k=1}\epsilon_{ijk}u_iv_jw_k\bigg)$$
But I am very weak with manipulations of sums and can't get much further. Any hints?
Note This problem occured when the text book formalized the corkscrew rule and discussed orientation of vectors in $\mathbb{R}^3$, meaning even matrices has not been introduced yet, as the chapter itself was analyzing vectors in the given space.
 A: First note that if $\bf{a},\bf{b},\bf{c}$ are not linearly independent,
then both sides are zero and we are done.
So we may assume that $(\bf{a},\bf{b},\bf{c})$ is a basis of ${\mathbb R}^3$. For the same reason we may further assume that $(\bf{u},\bf{v},\bf{w})$ is a basis of ${\mathbb R}^3$.Say that a sextuple of vectors $(\bf{a},\bf{b},\bf{c},\bf{u},\bf{v},\bf{w})$ is $\bf good$ if the desired equality holds.
Note that if we multiply one of the vectors in the sextuplet by a constant $\lambda$, then both sides of the equality are multiplied by $\lambda$, so this transformation does not affect goodness.
Similarly, if we make interchange $a$ and $b$, or interchange $u$ and $v$ (but do not
mix up any of $a,b,c$ with any of $u,v,w$), then both sides of the equality are unchanged,
so this transformation does not affect goodness.
Also, if we replace $b$ with $b+a$, or replace $v$ with $v+u$ (but do not
mix up any of $a,b,c$ with any of $u,v,w$), then both sides of the equality are unchanged,
so this transformation does not affect goodness.
Now it is well-known that iterating those transformations allows us to go from
the canonical  basis $(e_1,e_2,e_3)$ of ${\mathbb R}^3$ to any basis, and vice versa (those transformations are all revertible); this is the Gauss method to solve linear systems.
So we can assume without loss of generality that both $(\bf{a},\bf{b},\bf{c})$ and
$(\bf{u},\bf{v},\bf{w})$ are in fact the canonical basis, in which case both sides 
of the equality are clearly equal to $1$. This concludes the (computation-free) proof.
