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
$$
\frac{{\bf B}\times{\bf C}}{{\bf A}\cdot({\bf B}\times{\bf C})}
=
\frac{{\bf c}\cdot({\bf a}\times {\bf b})\,{\bf a}}{{\bf c}\cdot({\bf a}\times {\bf b})}
= {\bf a}
$$
