Product of two 3x3 determinants is the determinant of dot products this is a question from Alan Beardon's book "Algebra and Geometry."
Some notation,
$$[a, b, c] := a \cdot (b * c)$$
This is the determinant of 3 vectors in $\mathbb{R}^{3}$. note that $[a, b, c]$ is invariant under cyclic permutation so $[a, b, c] = [b, c, a] = [c, a, b].$ My job is to show that $$[a, b, c][u, v, w] = det((a \cdot u, a \cdot v, a \cdot w), (b \cdot u, b \cdot v, b \cdot w), (c \cdot u, c \cdot v, c \cdot w)).$$
I begin from the right side of the equation. By the rules of 3x3 determinants,
$$det((a \cdot u, a \cdot v, a \cdot w), (b \cdot u, b \cdot v, b \cdot w), (c \cdot u, c \cdot v, c \cdot w)) = $$ $$(a \cdot u)((b \cdot v)(c \cdot w) - (b \cdot w)(c \cdot v)) + (a \cdot v)((b \cdot w)(c \cdot u) - (b \cdot u)(c \cdot v)) + $$ $$ (a \cdot w)((b \cdot u)(c \cdot v) - (b \cdot v)(c \cdot u)).$$
Factoring out $b,$
$$(a \cdot u)(b \cdot ((c \cdot w)v - (c \cdot v)w)) + (a \cdot v)(b \cdot ((c \cdot u)w - (c \cdot w)u)) + $$ $$(a \cdot w)(b \cdot ((c \cdot v)u - (c \cdot u)v)).$$
With a little experience in vector manipulation we notice that the terms $b$ inner products with can be rewritten as cross products and we get the new form
$$(a \cdot u)(b \cdot (c * (v * w))) + (a \cdot v)(b \cdot (c * (w * u))) + (a \cdot w)(b \cdot (c * (u * v))).$$ I don't quite see how to proceed from here, so I reformulated the above statement the equivalent
$$b \cdot (c * ((v * w)(a \cdot u))) + b \cdot (c * ((w * u)(a \cdot v))) + b \cdot (c * ((u * v)(a \cdot w))) = $$ $$[b, c, (v * w)(a \cdot u)] + [b, c, (w * a)(a \cdot v)] + [b, c, (u * v)(a \cdot w)].$$
I tried to think how to apply linearity since the $[a, b, c]$ is linear in each vector...
Thanks in advance.
 A: $\def\d{\delta}$An approach using the Levi-Civita symbol: 
\begin{align*}
[a,b,c][u,v,w] &= a_i b_j c_k \varepsilon_{ijk} u_l v_m w_n \varepsilon_{lmn} \\
&= a_i b_j c_k u_l v_m w_n 
\left|\begin{array}{ccc}
\d_{il} & 
    \d_{im} 
        & \d_{in} \\
\d_{jl} 
    & \d_{jm} 
        & \d_{jn} \\
\d_{kl}
    & \d_{km}
        & \d_{kn} \\
\end{array}\right| \\
&= a_i b_j c_k u_l v_m w_n \d_{il}(\d_{jm}\d_{kn}-\d_{jn}\d_{km}) + \ldots \\
&= a\cdot u(b\cdot v c\cdot w - b\cdot w c\cdot v) + \ldots
\end{align*}
A: Suppose there is some orthonormal basis $\{\ell, m, n\}$.  Let $A$ be a linear map such that, for any $\theta \in \mathbb R^3$,
$$A(\theta) = (\theta \cdot a) \ell + (\theta \cdot b) m + (\theta \cdot c) n$$
See that $A$ can be represented as a matrix $\bar A$ on the basis $\{l, m, n\}$, and that $\det \bar A = [a, b, c]$.
Similarly, let $U$ be a linear map such that
$$U(\theta) = (\theta \cdot \ell) u + (\theta \cdot m) v + (\theta \cdot n) w$$
And also see that $U$ can be represented as a matrix $\bar U$ with $\det \bar U = [u, v, w]$.  (Hint: this is easier to prove for $\bar U$ than $\bar A$.  If you have difficulty proving it for $\bar A$, try $\bar A^T$ instead, which is more similar in structure to $\bar U$.)
Now, consider the map $M(\theta) = A(U(\theta))$.  The quantity $\det \bar M$ on the $\{\ell, m, n\}$ basis has the same form as the right-hand side as your proposition (within a transpose, which doesn't affect the determinant).
Finally, $\bar M = \bar A \bar U$, and we know that $\det \bar M = \det \bar A \det \bar U = [a, b, c][u,v,w]$, which is the left-hand side of your proposition.  Thus, the proposition is proven.
