= Solution
{c}
The <vector triple product identity> gives
$$
\mathbf b'\times\mathbf c'
=\frac{(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b)}{V^2}
=\frac{V\mathbf a}{V^2}=\frac{\mathbf a}{V}.
$$
Eliminate $V$ with <1/d/iii> to obtain the <reciprocal basis involution>:
$$
\boxed{\mathbf a=\frac{\mathbf b'\times\mathbf c'}{\mathbf a'\cdot(\mathbf b'\times\mathbf c')}.}
$$
This formula actually holds without the unit-length assumptions. Under those assumptions it also says $|\mathbf b'\times\mathbf c'|=1/|V|$; normalizing this <cross product> alone would lose the sign of $V$ and could return $-\mathbf a$.
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