reciprocal vectors I don't understand some of the terminology in this  question. I googled reciprocal vectors and got an article on reciprocal lattices, but I'm not sure if that is what they are talking about in this question. Also, when they say that ${\bf A}$, ${\bf B}$, and ${\bf C}$ are defined by ... plus cyclic permutations, again I looked at the wikipedia article on the subject, but I still do not understand the concept. Does anyone have a link for a clear explanation?

The vectors ${\bf a}$, ${\bf b}$, and ${\bf c}$ are non-coplanar, and
  form a non-orthogonal vector base. The vectors ${\bf A}$, ${\bf B}$,
  and ${\bf C}$, defined by 
$$ {\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot{\bf b}\times
 {\bf c}}, $$
plus cyclic permutations, are said to be reciprocal vectors. Show that
$$ {\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot{\bf B}\times {\bf
 C}}, $$
plus cyclic permutations.

thanks
 A: I think they intended to write $A\cdot (B\times C)$, which is a scalar; the vectors are  non-coplanar, hence this scalar is nonzero and hence we can safely divide by this scalar.
This wiki article would be a good start.
A: Cyclic means a cyclic permutation of the operands:
$$
{\bf A} \to {\bf B}, {\bf B} \to {\bf C}, {\bf C} \to {\bf A} 
\quad 
a \to b, b \to c, c \to a
$$
This gives
$$ 
{\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot({\bf b}\times {\bf c})}
\quad
{\bf B} = \frac{{\bf c}\times {\bf a}}{{\bf b}\cdot({\bf c}\times {\bf a})}
\quad
{\bf C} = \frac{{\bf a}\times {\bf b}}{{\bf c}\cdot({\bf a}\times {\bf b})} 
$$
and 
$$ 
{\bf a} = \frac{{\bf B}\times {\bf C}}{{\bf A}\cdot({\bf B}\times {\bf C})}
\quad
{\bf b} = \frac{{\bf C}\times {\bf A}}{{\bf B}\cdot({\bf C}\times {\bf A})}
\quad
{\bf c} = \frac{{\bf A}\times {\bf B}}{{\bf C}\cdot({\bf A}\times {\bf B})}
$$
From Reciprocal Lattice

[..] the "crystallographer's" definition, comes
  from defining the reciprocal lattice to be $e^{2 \pi
 i\mathbf{K}\cdot\mathbf{R}}=1$ which changes the definitions of
  the reciprocal lattice vectors to be
$ \mathbf{b_{1}}=\frac{\mathbf{a_{2}} \times
 \mathbf{a_{3}}}{\mathbf{a_{1}} \cdot (\mathbf{a_{2}} \times
 \mathbf{a_{3}})}$ and so on for the other vectors.  The
  crystallographer's definition has the advantage that the definition of
  $\mathbf{b_{1}}$ is just the reciprocal magnitude of
  $\mathbf{a_{1}}$ in the direction of $\mathbf{a_{2}}
 \times \mathbf{a_{3}}$, dropping the factor of $2
 \pi$.

Checking the magnitude:
$$ 
\lVert{\bf A}\rVert = \frac{\lVert {\bf b} \times{\bf c}\rVert}{\lVert{\bf a}\rVert\lVert{\bf b}\times {\bf c}\rVert\cos\angle({\bf a}, {\bf b} \times{\bf c})}
=
\frac{1}{\lVert{\bf a}\rVert ({\bf e_a} \cdot {\bf e}_{{\bf b} \times{\bf c}})} 
$$
Solving the question:
Using the "bac-cab" rule 
${\bf a}\times({\bf b}\times{\bf c}) = {\bf b}({\bf a}\cdot {\bf c}) - {\bf c}({\bf a}\cdot {\bf b})$ we go for the nominator of ${\bf B}\times{\bf C}$:
$$
({\bf c}\times {\bf a})\times({\bf a}\times{\bf b}) =
{\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})-{\bf b}(({\bf c}\times{\bf a})\cdot{\bf a})=
{\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})
$$
because ${\bf c}\times{\bf a} \perp {\bf a}$.
This gives
$$
{\bf B}\times{\bf C} 
= 
\frac{{\bf a}(({\bf c}\times{\bf a})\cdot{\bf b})}{({\bf b}\cdot({\bf c}\times {\bf a}))({\bf c}\cdot({\bf a}\times {\bf b}))}
= 
\frac{{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})} \\
{\bf A}\cdot({\bf B}\times{\bf C})
=
\frac{{\bf b}\times {\bf c}}{{\bf a}\cdot({\bf b}\times {\bf c})}
\cdot
\frac{{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})}
=
\frac{1}{{\bf c}\cdot({\bf a}\times {\bf b})}
$$
Dividing those gives the desired result
$$
\frac{{\bf B}\times{\bf C}}{{\bf A}\cdot({\bf B}\times{\bf C})}
=
\frac{{\bf c}\cdot({\bf a}\times {\bf b})\,{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})}
= {\bf a}
$$
