= Solution
For the $n$ distinct complex embeddings $\sigma_1,\ldots,\sigma_n$ of the <number field>, define the <discriminant of elements of a number field> by
$$
D(w_1,\ldots,w_n)=\det(\sigma_i(w_j))^2
=\det\bigl(\operatorname{Tr}_{K/\mathbb Q}(w_iw_j)\bigr).
$$
The equality follows by multiplying the transpose of the embedding <matrix> by itself. Each <field trace> is rational: it is also the trace of the rational <matrix> of multiplication by $w_iw_j$ on $K$. Thus the <determinant> lies in $\mathbb Q$. If the $w_i$ are <algebraic integers>, the products and their conjugates are <algebraic integers>, and so are their sums. A rational <algebraic integer> is an <integer>, hence all trace entries and their <determinant> are <integers>.
Write $u_j=\sum_i a_{ij}v_i$. The inclusion of the generated abelian <groups> makes $A=(a_{ij})$ an <integer> <matrix>. Its embedding <matrix> is the embedding <matrix> of the $v_i$ multiplied by $A$. Taking squared <determinants> gives
$$
\boxed{D(u_1,\ldots,u_n)=(\det A)^2D(v_1,\ldots,v_n).}
$$
For independent <algebraic integers> $v_i$, choose an <integral basis> $\omega_i$ of $\mathcal O_K$. The same formula gives $D(v)=[\mathcal O_K:\sum_i\mathbb Zv_i]^2D(\omega)$. If $D(v)$ is nonzero and a <square-free integer>, the <integer> index must be one, proving that the $v_i$ form an <integral basis>.
Now put $\beta=3\alpha+4$. Since $\mathbb Q(\beta)=\mathbb Q(\alpha)$ has degree three, the <monic polynomial>
$$
27f\!\left(\frac{X-4}{3}\right)=X^3-6X^2+59
$$
is its <minimal polynomial of an algebraic element>. The <field norm> is the negative of its constant term:
$$
\boxed{\operatorname{Norm}_{K/\mathbb Q}(3\alpha+4)=-59.}
$$
For the discriminant calculation, $f'(\alpha)=3\alpha^2+4\alpha$. In the <basis> $1,\alpha,\alpha^2$, using $\alpha^3=-2\alpha^2-1$, multiplication by this element has <matrix>
$$
M=\begin{pmatrix}0&-3&2\\4&0&-3\\3&-2&4\end{pmatrix},\qquad\det M=59.
$$
Thus the given derivative-norm formula gives $D(1,\alpha,\alpha^2)=-59$. Since $\alpha$ satisfies a <monic polynomial> with <integer> coefficients, $1,\alpha,\alpha^2$ are <algebraic integers>; the fact that a <squarefree discriminant certifies an integral basis> proves
$$
\boxed{\mathcal O_K=\mathbb Z[\alpha],\qquad D_K=-59.}
$$
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