Across the cut $A|BC$, the two conditional $BC$ <vectors> $\alpha|00\rangle$ and $\gamma|00\rangle+\beta|11\rangle$ are linearly independent when $\alpha\beta\ne0$, so the Schmidt rank exceeds one. Hence the state cannot be a three-qubit product for any $\gamma$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 15E / a / Solution

 Solution (source code)