Solution
= Solution
{c}
Apply the <squared scalar triple product from cyclic cross products> proved in <1/c>:
$$
\mathbf a'\cdot(\mathbf b'\times\mathbf c')
=\frac{(\mathbf b\times\mathbf c)\cdot[(\mathbf c\times\mathbf a)\times(\mathbf a\times\mathbf b)]}{V^3}
=\frac{V^2}{V^3}.
$$
Hence the <reciprocal basis> has <scalar triple product> $\boxed{\mathbf a'\cdot(\mathbf b'\times\mathbf c')=1/V}$, with its sign retained.