I’m going through Marcus number Field chapter 3 an I’m finding very hard to understand the part about the decomposition of pR (theorem 27) that tells us that if $pnotR/Z[alpha ]$ then we can decompose $pR$ by looking at a factorization of it’s minimal polynomial (Kummer’s theorem?)
In partcular I’m stuck on exercise 26
Let $alpha={ ^{3}sqrt{m}}$ where m is a cubefree integer, $K = mathbb{Q}[alpha]$, $R = mathbb{A} cap mathbb{Q}[alpha]$
Show that if p is a prime $neq 3$ and $p^2 notm$ , then the prime decomposition of pR
can be determined by factoring $x^3 − m; mod; p.$ (See Theorem 27 and exercise
41, chapter 2 (this tells us the discriminat and the integral bases I write below).)Suppose $p^2  m$. Writing $m = hk^2$ as in exercise 41, chapter 2, set $ gamma= frac{alpha^2}{k}.$
Show that p does not divide $R/Z[gamma ]$; use this to determine the prime decomposition
of pR.Determine the prime decomposition of 3R when $mnotequiv pm 1$ (mod 9).
 Determine the prime decomposition of 3R when m = 10. (Hint: Set $beta = (alpha −
1)^2/3$ and use exercise 18 to show that disc(β) = 4 disc(R). Also note exercise
41(d), chapter 2 (this tells us that $beta^3beta^2+left(frac{
1+2m}{3}right)betafrac{(m1)^2}{27}=0))$ Show that this always works for $mequiv pm 1; (mod; 9)$ except
possibly when $mequiv pm 8; (mod; 27)$. Show that 9 $not$ disc(R) when $mequiv pm 1; (mod; 9)$; use this to show that 3R is not
the cube of a prime ideal. Assuming the converse of Theorem
24, show that 3R = $P^2Q$ where P and Q are distinct primes of R.
I think I’ve done point 1) using the fact that $p^2not disc(alpha)$ implies we can use theorem 27 that tells us exactly that we can decompose pR simply by factorizing the minimal polynomial of $alpha$, but the problem is now point 2) (and the ones after since they rely on 2).
I was able to prove that $gamma=sqrt{h^2k}$ and that $p^2not h^2k$ so either we can use the fact above or $p^2disc(alpha)=27^2*k^2hRightarrow p^227^2$ so p=3, but now I don’t know how to prove that 3 doesn’t divide $R/mathbb{Z}[alpha]$ since for me the latter is always divisible by 3.
An integral base of the above is either
$$left(1,alpha,frac{alpha^2+k^2alpha+k^2}{3k}right),quad left(1,alpha,frac{alpha^2k^2alpha+k^2}{3k}right),quad left(1,alpha,frac{alpha^2}{k}right) $$
if respectivly $mequiv 1; (mod; 9)$, $mequiv 1; (mod; 9)$, $mnotequiv pm1; (mod; 9)$
Any help would be welcomed, even more if quite specific on the calculations since I think there is something I miss on a theoretical level.
Exercise 18 Let K be a number field of degree n over $mathbb{Q}$ , and let $alpha_1, dots , alpha_n in K.$
Show that $disc(ralpha_1, alpha_2, dots , alpha_n) = r^2 disc(alpha_1, dots , alpha_n)$ for all r $in mathbb{Q}$.
Let $beta$ be a linear combination of $alpha_2, dots , alpha_n$ with coefficients in $mathbb{Q}$. Show that
$disc(alpha_1 + beta, alpha_2, dots , alpha_n) = disc(alpha_1, dots , alpha_n).$Theorem 24 Let p be a prime in $mathbb{Z}$, and suppose p is ramified in a number ring R.
Then p  disc(R).
UPDATE: The question is still without answer so for now I’ll post my solution to the first two points, then if a better one comes I’ll be happy to set it as solving the question.

Uniforming the notation between this exercise and therem 27 of Marcus we have
$$ L=mathbb{Q}[alpha]quad S=mathbb{A}cap mathbb{Q}[alpha]quad K=mathbb{Q}quad R=mathbb{Z}$$ so to use theorem 27 we have to check
$$ pnotleftfrac{mathbb{A}cap mathbb{Q}[alpha]}{mathbb{Z}[alpha]}right$$ but actually first we can use the corollary telling us that the hypotesis are satisfied if $p^2notdisc(alpha)$,
exercise 41 in chapter 2 tells us that in our case $disc(alpha)=27^2m$ and so if $pneq 3wedge p^2m$ we are in the hypotesis of the corollary and thus of the theorem and so we can decompose pR by factoring $x^3m$; 
In this case the hypotesis of the corollary are not satisfied. We have also that $p^2miff p^2hvee p^2k^2$ but since h is squarefree we have that it has to be $p^2k^2iff pkiff pnoth$ since they are coprime. Now we can write
$$ alpha=sqrt{hk^2}iff alpha^2=sqrt{h^2k^4}iff gamma=frac{alpha^2}{k}=sqrt{h^2k}$$ and we have that $p^2h^2kiff p^2h^2iff ph$ which is not true so $p^2not h^2k$.
But now $ph^2k=n$ but $p^2noth^2k$ so $x^3n$ is a pEisentstein polynomial and we can use the following theorem to deduce $$pnot mathbb{A}cap mathbb{Q}[gamma]/mathbb{Z}[gamma]$$
Let K = $mathbb{Q}(alpha)$ where $alphain mathbb{A}cap mathbb{Q}[alpha]$ is the root of an Eisenstein polynomial at p,
with degree n. Then $p not mathbb{A}cap mathbb{Q}[alpha] / mathbb{Z}[alpha]$.