= Solution
Since $p/q$ is not a root, $q^df(p/q)$ is a nonzero <integer>, and hence
$$
|f(p/q)|\geq q^{-d}.
$$
Moreover $0<\theta-p/q<1$ implies $|p/q|<|\theta|+1$. Applying part (i) to $s=\theta$, $t=p/q$, and using $f(\theta)=0$ gives
$$
q^{-d}\leq A(|\theta|+1)^{d-1}\left|\theta-\frac pq\right|.
$$
Rearrangement proves
$$
\boxed{\theta-\frac pq\geq
\frac1A\left(\frac1{|\theta|+1}\right)^{d-1}\frac1{q^d}}.
$$
This is the core estimate behind the <Liouville approximation theorem>.
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