I am trying to learn basics from the book on Clifford Algebras and Spinors 2nd ed by P. Lounesto and I can't get past the last equation on page 43;$x \wedge y \wedge z$. Is that a derived expression? I understand the one before, $x \wedge y$, which is easily obtained, so I'm expecting the following one can be obtained as well, but I don't see how. On the following page Lounesto states that there is another construction, for $x \wedge y \wedge z$, which sounds to me as a definition. These two constructions are different which is confusing me. The second one is related to $x \wedge B$, but how it's still a puzzle to me. If someone can provide some guidance, I'd appreciate it.
UPDATE: I apologize. I thought that anybody who knows this stuff would have the book. Here are the equations:
$x \wedge y=\frac{1}{2}(xy-yx)$ $\qquad$ (a);
$x \wedge y \wedge z =\frac{1}{6} (xyz+yzx+zxy-zyx-xzy-yxz)$ $\qquad$ (b);
$x \wedge y \wedge z =\frac{1}{2} (xyz-zyx)$ $\qquad$ (c);
$x \wedge B=\frac{1}{2}(xB+Bx)$ $\qquad$(d);
where $x,y,z$ are vectors, and $B$ is a bi-vector. So, equation (a) is easily obtained from the definition of the Clifford product, but Lounesto says the subspace of 3-vectors can be uniquely reconstructed within $Cl_3$ "by completely antysymmetrized Clifford product" and he gives equation (b) (how to derive this? Since I saw 6 in there, I tried permuting $x,y,z$ and adding up the equations, but it is not coming up). Then he says there is another construction of the subspace of 3-vectors, obtained using the reversion, equation (c), (is this a definition? where is this coming from?), related to the recursive construction equation(d) (how is it related?).
UPDATE 2: I was as able to derive both b) and c). b) first by using formula a) on all pairs of different vectors and multiplying it with remaining vectors, then by using $Ba=B \cdot a + B \wedge a$, and then applying usual formula $a \cdot (b \wedge c) = (a \cdot b)c -(a \cdot c)b$ to the left hand sides of equations before adding them up.
And I was able to get c) by using $a\wedge B = \frac{1}{2}(aB+Ba)$ and expanding the terms in the brackets on the right hand side and using $y \cdot z$ in terms of Clifford product to obtain $ -zyx$. So now, unless I made a mistake, I don't understand how b) and c) can be equal because b) already contains terms $ xyz -zyx $.
The answer to my last question is rather simple. Since c) can be demonstrated, by a simple renaming schemes the other two combinations, $yzx-xzy$ and $zxy-yxz$ can also be proved (these two combinations are also in b) ). So, 3 equations which have the form of c) can be added up to obtain b).