Question: Let $a>1$ be a cubefree integer and $p$ a prime with $p \nmid 3a$. Show $p$ splits completely in $\mathbb{Q}(\sqrt[3]{a})$ iff $a$ is a cube mod p and $p \equiv 1 \:\:\text{mod} \:3$. 

Really struggling with where to go with this one :/

Answer: 
Let $K=\mathbb{Q}(\sqrt[3]{a})$ and let $f=X^3-a$, which is irreducible since $a$ is cube free, so it is the minimal polynomial of $\sqrt[3]{a}$. 

Now $disc(f)=27a^2$. Since $disc(f)=[\mathcal{O}\_K:\mathbb{Z}[\sqrt[3]{a}]]^2d\_K$, (where $d\_K$ is the discriminant of $K$), a prime $p$ which does not divide $3a$ does not divide $[\mathcal{O}\_K:\mathbb{Z}[\sqrt[3]{a}]]$ either. 

By a famous theorem of Dedekind, the decomposition of $p$ in this case is reflected by the decomposition of $f$ modulo $p$. In particular, $p$ is totally split if and only if $f$ mod $p$ splits into three linear factors. 

Now, $f$ splits mod $p$ if and only if $a$ is a cube mod $p$ and $\mathbb{F}\_p$ has a 3rd root of $1$ (easy). Since $\mathbb{F}\_p^\times$ is cyclic, the last condition is easily seen to be equivalent to $3\mid p-1$.

Details are left to you ;-)
For the splits completely problem we can look at the reduction $\bmod p$, then $p$ splits completely in $\Bbb{Z}[\sqrt[3]{a}]$ iff $X^3-a$ has 3 distinct roots $\bmod p$ which happens when $a\equiv b^3\bmod p$ and $c^3\equiv 1\bmod p$ ie. $p\equiv 1\bmod 3$.
Thus $$(p)= (p,\sqrt[3]{a}-b)(p,\sqrt[3]{a}-bc)(p,\sqrt[3]{a}-bc^2)$$ as ideals of $\Bbb{Z}[\sqrt[3]{a}]$ thus as ideals of $O\_K$.
Conversely $p\nmid 3a$ implies that $X^3-a$ is separable $\bmod p$ then $(p)$ splits completely in $O\_K$ iff $(p)= (p,\sqrt[3]{a}-b)(p,\sqrt[3]{a}-bc)(p,\sqrt[3]{a}-bc^2)$ where $b,c$ as integers so the decomposition stays true in $\Bbb{Z}[\sqrt[3]{a}]$.

If the problem was to find when it is inert or non-inert non split then we'd need to look at finite extensions of $\Bbb{Q}\_p$.