= Solution
The element $\alpha$ is a root of
$$
f(T)=T^3-5T+8.
$$
By the <rational root theorem>, any rational root of this monic polynomial must be one of
$$
\mathord\pm1,\mathord\pm2,\mathord\pm4,\mathord\pm8.
$$
Direct substitution shows that none is a root. A reducible cubic over a field has a linear factor, so $f$ is an <irreducible polynomial> over $\mathbb Q$. It is therefore the <minimal polynomial of an algebraic element>[minimal polynomial] of $\alpha$, and
$$
\boxed{[K:\mathbb Q]=\deg f=3}.
$$
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