= Solution
The polynomial is an <Eisenstein polynomial> at $5$, so it is irreducible and $[K:\mathbb Q]=3$. Its <polynomial discriminant> is
$$
\operatorname{disc}(f)=500-10800=-10300=-4\cdot5^2\cdot103.
$$
One must remove the index at $2$ before using this as the <field discriminant>. Put $\beta=(\alpha^2+\alpha)/2$. Direct reduction using $\alpha^3=5\alpha-20$ gives
$$
\alpha^2=2\beta-\alpha,\quad \alpha\beta=2\alpha+\beta-10,\quad \beta^2=3\beta-4\alpha-10.
$$
Thus $B=\mathbb Z+\mathbb Z\alpha+\mathbb Z\beta$ is a ring finite over $\mathbb Z$, so its elements are integral. Its basis has <discriminant of elements of a number field> $-2575=-5^2\cdot103$. The <discriminant-index formula for an integral lattice> says $\operatorname{disc}(B)=[\mathcal O_K:B]^2\Delta_K$, so only $5$ could divide the remaining index.
We use these standard <different ideal> facts: the norm of the different is the absolute <field discriminant>; its localization is the local different; an unramified prime has different exponent zero, and a tamely ramified prime of <ramification index> $e$ has different exponent $e-1$. The <Eisenstein polynomial> at $5$ gives total ramification of degree three, which is tame because $5\nmid3$. Its contribution to $v_5(\Delta_K)$ is therefore two. This excludes a further index factor $5$, proving \b[$\mathcal O_K=B$ and $\boxed{\Delta_K=-2575}$.]
Since the index $[\mathcal O_K:\mathbb Z[\alpha]]=2$ is prime to $5$ and $103$, <Dedekind factorization theorem> applies at both primes. Modulo $5$, $f=X^3$, giving a single prime $\mathfrak p=(5,\alpha)$ with <residue degree> one and <ramification index> three. Modulo $103$,
$$
f(X)=(X-6)^2(X+12).
$$
Thus $\mathfrak q=(103,\alpha-6)$ has <residue degree> one and <ramification index> two; the other prime over $103$ is unramified. Both ramified primes are tame. No other prime ramifies because none divides $\Delta_K$. \b[Consequently]
$$
\boxed{\mathfrak D_{K/\mathbb Q}=\mathfrak p^2\mathfrak q,\qquad N\mathfrak p=5,\quad N\mathfrak q=103.}
$$
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