Factorization of an ideal Let $1< x\in \mathbb{Z}$ is not divisible by the cube of any integer $> 1$,  So the ﬁeld $K = \mathbb{Q}(\sqrt[3]{x})$, where $\sqrt[3]{x}$ is a cubic extension of $\mathbb{Q}$. 
Suppose $x$ be not divisible by the square of any integer $> 1$ and let $p$ be a prime number dividing $x$. Show that the factorization of the ideal $pOK = p^3$.
Suppose 3 does not divide $x$ and  $x$ is not equivalent to $±1$. 
Show then that the prime number 3 ramifies in K.
In general, I am struggling with square free and cube free.
 A: Taking completions generally works well for this kind of question. So consider the field $\mathbf Q_p (\alpha)$, where $\alpha$ is a root of your $x$. The condition put on $x$ can be interpreted as the existence of a rational prime $p$ s.t. $p \mid x$ and $p^2 \nmid x$. This means that our polynomial is of Eisenstein type, and Eisenstein 's criterion (see any textbook) asserts that it is irreducible over $\mathbf Q_p$, and that $\alpha$ is a uniformizer of $\mathbf Q_p (\alpha)$, i.e. the local extension $\mathbf Q_p (\alpha)/\mathbf Q_p$ is totally ramified of degree 3. Since the global degree verifies $3=\sum e_P.f_P$, with $e_P$=ramification index, $f_P$ = inertia index, the sum bearing on all the prime ideals $P$ above $p$, it follows that there is only one $P$ above $p$, with $e_P=3$. In other words, $(p)=P^3$.
In your second question, a modulus is missing. The congruence  seems   to be $x\neq\pm1$mod $3$, but then, since $\mathbf F_3 =(0, \pm 1)$ , there is a contradiction with the condition $3\nmid x$ . 
