Prove that $x^3-2$ and $x^3-3$ are irreducible over $\Bbb{Q}(i)$ 
Let $F=\Bbb{Q}(i)$. Prove that $x^3-2$ and $x^3-3$ are irreducible over $F$.

How do I go about this? Should I just say that the roots of $x^3-2$ are $2^{1/3},2^{1/3}e^{i2\pi/3},2^{1/3}e^{i4\pi/3}$, which do not belong to $\Bbb{Q}(i)$?
 A: A lot of the other answers have espoused that your answer is ultimately ok, but you should be cautious with polynomials of higher degree. I can't say I fully agree with the first point - saying that the roots aren't in $\mathbb{Q}(i)$ feels to me like you are begging the question, because that is precisely what you are trying to prove. How do you know that $2^{1/3}$ is not in $\mathbb{Q}(i)$, other than because it seems like it should be true?
To me, there are two rigorous approaches that stick out here. The first is the Eisenstein argument given in D_S's answer, which is great and very beautiful. It also hinges on knowing a few (albeit elementary) facts about $\mathbb{Z}[i]$, which you may not feel comfortable using.
The other argument that comes to my mind is a degree argument. This, I believe, proves what you are trying to show is true. Let $\alpha$ be a root of $f(X) = X^{3}-2$. If $f$ is reducible over $\mathbb{Q}(i)$, then without loss of generality, $\alpha$ is in $\mathbb{Q}(i)$, so $\mathbb{Q}(\alpha)$ is a subfield of $\mathbb{Q}(i)$. By the multiplicativity of field extension degree, $[\mathbb{Q}(i):\mathbb{Q}] = [\mathbb{Q}(i):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}] = 2$, so $[\mathbb{Q}(\alpha):\mathbb{Q}]\mid 2$. But by Eisenstein at $2$, $f$ is irreducible over $\mathbb{Q}$, so $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$, a contradiction. An identical argument works for $X^{3}-3$.  
A: If a cubic is reducible then one of the factors is linear, which means it should have a root in $\mathbb{Q}(i)$. So your method is correct. Notice that a factor of a quartic may be quadratic so simply showing it has no roots isn't enough.
A: For a cubic polynomial, that argument works: it's irreducible if and only if it doesn't have any roots.
Alternately for $X^3 - 3$, you can use the fact that $\mathbb{Z}[i]$ is a unique factorization domain and $3$ is prime in $\mathbb{Z}[i]$, and apply the Eisenstein criterion.
A: Almost, except that the correspondence $$P(x) \text{ doesn't have roots }\leftrightarrow P(x) \text{ is irreducible }$$
Is false for an arbitrary polynomial over an arbitrary field, and false in your specific field, $\Bbb Q(i)$, as well (for example, $(x^2-2)^2$ has no roots but is clearly not irreducible).
However, the argument works for these polynomials, since their degree is $3$ (and therefore...?).
A: Here is an idea. 
Notice that, by Eiseinstein, $X^3 - 2 $ is irreducible over $\mathbb Q$, and it has degree $3$. Consider the polynomial  $X^2 + 1$ responsible for the extension $\mathbb Q [i]$ over $\mathbb Q$, and take the extension $Gal (\mathbb Q[\sqrt 2 , i];\mathbb Q)$ we may conclude from the fact that $\gcd \{2,3\} = 1$ that $$[\mathbb Q[\sqrt 2 , i]:\mathbb Q[i]] = 3 $$ and $X^3-2$ is irreducible over $\mathbb Q[i]$. 
A: If $x^3-2$ had a root $\xi$, we would have:
$$N(2)=4=N(\xi)^3.$$
However $4\,$ is not a cube in $\mathbf Z$. Same argument for $x^2-3$.
