For $\beta\in K$, the nonzero algebraic number $\gamma=\alpha-\beta$ has degree at most
$$
[\mathbb Q(\alpha,\beta):\mathbb Q]\leq3\cdot2=6.
$$
The basic height inequality gives
$$
H(\gamma)\leq2H(\alpha)H(\beta).
$$
The <Liouville height inequality> therefore yields
$$
|\alpha-\beta|
\geq H(\gamma)^{-6}
\geq(2H(\alpha))^{-6}H(\beta)^{-6}.
$$
Thus one may take the explicit constant
$$
c=(2H(\alpha))^{-6}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 166 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-166/3/a/solution