= Solution
The <Dedekind factorization theorem> says: if $K=\mathbb Q(\theta)$ with $\theta$ integral, <minimal polynomial of an algebraic element> $f\in\mathbb Z[X]$, and the rational prime $p$ does not divide $[\mathcal O_K:\mathbb Z[\theta]]$, factor $\bar f=\prod_i\bar f_i^{e_i}$ into distinct monic irreducibles over $\mathbb F_p$. Then the distinct <prime ideals> above $p$ are $P_i=(p,f_i(\theta))$, with <integer> lifts $f_i$, and
$$
\boxed{(p)=\prod_iP_i^{e_i},\qquad N(P_i)=p^{\deg\bar f_i}.}
$$
The index condition is part of the theorem and must be checked before applying it at a ramified prime.
For this cubic, $X^3-6$ is Eisenstein and has <discriminant> $-972$. We first verify $\mathcal O_K=\mathbb Z[\theta]$. The square of the index divides 972, so only two and three can divide the index. If either prime $p$ divided it, the finite quotient <group> would contain an element of order $p$, giving an integral element $(a+b\theta+c\theta^2)/p$ with digits $0\le a,b,c<p$, not all zero. Its <field norm> is
$$
\frac{a^3+6b^3+36c^3-18abc}{p^3}.
$$
For $p=2$, the numerator has valuation zero if $a=1$, valuation one if $a=0,b=1$, and valuation two if $a=b=0,c=1$. For $p=3$ the same three cases, according to the first nonzero digit, give valuations zero, one and two. None gives an <integer> <field norm> after dividing by $p^3$, contradicting integrality. Thus the index is one.
Multiplication by $a+b\theta$ in the <power basis>, or the product over the three conjugates, gives $\boxed{N(a+b\theta)=a^3+6b^3}$. Define
$$
\begin{aligned}
P_2&=(2,\theta),&P_3&=(3,\theta),\\
P_5&=(5,\theta-1),&Q_5&=(5,\theta^2+\theta+1),\\
P_{7,r}&=(7,\theta-r)&& (r=3,5,6).
\end{aligned}
$$
Reduction of the cubic gives $X^3$ at two and three, $(X-1)(X^2+X+1)$ at five, and $(X-3)(X-5)(X-6)$ at seven. The quadratic factor at five is irreducible since its <discriminant> is two, a nonsquare modulo five. Hence
$$
\boxed{(2)=P_2^3,\quad(3)=P_3^3,\quad(5)=P_5Q_5,\quad(7)=P_{7,3}P_{7,5}P_{7,6}.}
$$
The <norms> are respectively $2,3,5,25,7,7,7$. <Norms> and reduction of the elements at those primes give
$$
\boxed{\begin{aligned}
(\theta)&=P_2P_3,&(\theta-1)&=P_5,&(\theta-2)&=P_2,\\
(\theta-3)&=P_3P_{7,3},& (\theta+4)&=P_2P_5P_{7,3}.
\end{aligned}}
$$
For example their absolute element <field norms> are $6,5,2,21,70$; each indicated prime divides the element by substitution in its <residue field>, and the product of the indicated <ideal norms> already equals that absolute <field norm>. Thus no additional factor or exponent is possible.
In the <ideal> of the specified quotient every prime exponent cancels:
$$
(\mu)=\frac{P_2^2(P_3P_{7,3})P_5}{(P_2P_3)(P_2P_5P_{7,3})}=\mathcal O_K.
$$
Equality with the <unit ideal> implies that both $\mu$ and its inverse are integral, so $\boxed{\mu\text{ is a unit}}$. For the real root, $1<\theta<2$. The numerator is negative and has absolute value less than two, while the denominator is greater than five. Thus $-2/5<\mu<0$, proving $\boxed{\mu\ne\pm1}$. Its <field norm> is $-1$, consistent with the <ideal> calculation.
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