= Solution
Put
$$
\beta=\frac{\alpha+\alpha^2}{2}.
$$
Using $\alpha^3=5\alpha-8$, multiplication by $\beta$ on the ordered basis $(1,\alpha,\beta)$ is determined by
$$
\beta\cdot1=\beta,
\qquad
\beta\alpha=-4+2\alpha+\beta,
\qquad
\beta^2=-4-\alpha+3\beta.
$$
Its matrix, with these coordinate vectors as columns, is
$$
M_\beta=
\begin{pmatrix}
0&-4&-4\\
0&2&-1\\
1&1&3
\end{pmatrix}.
$$
The <characteristic polynomial> is
$$
\det(TI-M_\beta)=T^3-5T^2+11T-12.
$$
By the <Cayley-Hamilton theorem>, this monic integer polynomial annihilates $\beta$, so $\beta$ is an <algebraic integer>.
Let
$$
A=\mathbb Z\langle1,\alpha,\beta\rangle.
$$
The element $\alpha$ is also integral because it satisfies the monic polynomial $f$. Hence $A\subseteq\mathcal O_K$. Since
$$
\alpha^2=2\beta-\alpha,
$$
the lattice $\mathbb Z[\alpha]$ has index two in $A$. The given power-basis discriminant is
$$
\Delta(1,\alpha,\alpha^2)=-4\cdot307,
$$
so the <discriminant-index formula for an integral lattice> gives
$$
\Delta(1,\alpha,\beta)
=\frac{-4\cdot307}{2^2}
=-307.
$$
If $m=[\mathcal O_K:A]$, then
$$
-307=m^2d_K.
$$
Because $307$ is prime, the square $m^2$ can divide $307$ only when $m=1$. Thus
$$
\boxed{\mathcal O_K=A
=\mathbb Z\langle1,\alpha,\beta\rangle},
$$
so $(1,\alpha,\beta)$ is an <integral basis of the cubic field of discriminant minus 307>.
Finally, the relation $\beta^2=-4-\alpha+3\beta$ gives
$$
\alpha=-4+3\beta-\beta^2.
$$
\b[Consequently $\mathcal O_K=\mathbb Z[\beta]$, which will allow direct use of Dedekind's theorem.]
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