= Solution
Expand a normalized <product state> as
$$
(a|0\rangle+b|1\rangle)\otimes(c|0\rangle+d|1\rangle)
=ac|00\rangle+ad|01\rangle+bc|10\rangle+bd|11\rangle.
$$
Thus $\alpha=ac$, $\beta=ad$, $\gamma=bc$, $\delta=bd$, giving
$$
\boxed{\alpha\delta=abcd=\beta\gamma.}
$$
Conversely the nonzero coefficient matrix $C=\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$ has determinant zero precisely when it has rank one, so the condition also suffices for a pure <two-qubit state> to factor.
For any fixed pair of local measurement settings with operators $E_i,F_j$, the <Born rule> factorizes:
$$
\boxed{p(i,j)=\langle a|E_i|a\rangle\langle b|F_j|b\rangle=p_A(i)p_B(j).}
$$
Hence a <product state> has independent local results in every pair of local bases. This assumes independent settings and local measurements; shared classical randomization of preparations or a subsequent joint interaction can generate correlations. A separable mixture is not necessarily a single <product state>.
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