Quarky Volume Formula for Parallelepiped

Beginning with the ubiquitous triple product identity

               Ax(BxC) = B(A.C)-C(A.B)                       (1)

for vectors in R^3, we immediately have the handy formula for the 
cross product of two cross products

        (AxB)x(CxD)  =  B[(CxD).A]  -  A[(CxD).B]            (2)

Now suppose we define three new vectors in terms of A,B,C as follows

       A' = B x C       B' = C x A        C' = A x B

It's well known that the triple product A' . (B' x C') is simply 
the square of the triple product A . (B x C).  To see why this is 
true, notice that using (2) the "primed" triple-product can be 
expressed as

     A'.(B' x C') =  (BxC) . ((CxA)x(AxB))

                  =  (BxC) . { A[(AxB).C] - C[(AxB).A] }

Since the quantity (AxB).C equals the volume of the parallelepiped 
with the vectors A,B,C forming adjacent edges, it's clear that 
(AxB).A = 0, because two of the three edges are identical.  Thus the 
right-most term in the preceeding expression drops out, and the
resulting quantity is simply (BxC).A[(AxB).C], which, in view of 
associativity of scalar multiplication over the dot product, is the 
same as [(BxC).A][(AxB).C].  Both factors equal the volume of a 
parallelepiped with edges A,B,C, so the product equals [A.(BxC)]^2.

But this is not very elegant.  Of course, in the special case 
when A,B,C are mutually orthogonal it follows that the pairwise
perpindicular vectors A',B',C' are also orthogonal, and so the
two triple products are just the volumes of two rectangular solids,
one with sides A,B,C and the other with sides A'=BC, B'=AC, C'=AB,
from which it's obvious that the second volume is the square of
the first.  This is based on the volume formula V=|A||B||C| for
rectangular solids, and the same degree of obviousness could be 
achieved in the general case if we had a suitable formula for the
volume of a general parallelpiped.

The triple product A.(BxC) corresponds to the volume of a general
parallelepiped with adjacent edges A,B,C, and this volume can be
expressed by the formula

            V  =  |A||B||C|  S^(2/3)  S'^(1/3)

where S is the product of the sines of the pairwise angles between
the vectors A,B,C, and S' is the product of sines of the dual vectors 
A',B',C'.  (This formula can be verified by substituting the vector 
expressions for S and S'.)  Likewise the volume of the dual parallel-
epiped is
           V' =  |A'||B'||C'|  S'^(2/3)  S^(1/3)

The magnitudes of the primed vectors are |B||C|sin(B_C),
 |A||C|sin(A_C), and |A||B|sin(A_B), so we have

       V'  =  (|A||B||C|)^2  S'^(2/3)  S^(4/3)   =  V^2

It's interesting how the 2/3 and 1/3 exponents on S and S' seem
to mimic the combinatorial behavior of the properties of quarks.
It's also interesting that S^(1/3) is just the geometric mean of
the three pairwise sines, so if we imagined a rectangular solid with
edge lengths equal to the three sines, the quantity S^(1/3) would
be the edge of a cube with the same volume.