Show that $x^{3}-3$ irreducible over $\mathbb{Q}(\sqrt{-3})$

There's a number of ways of attacking this problem. Here's but one way of going about this that requires a little less calculation:

Since $\deg(f) \leq 3$, we know $f(x) = x^3 - 3$ is reducible in $\mathbb{Q}(\sqrt{-3}) \iff$ a root of $f$ is contained in $\mathbb{Q}(\sqrt{-3})$.

Let $\alpha$ be a root of $f$. If $\alpha \in \mathbb{Q}(\sqrt{-3})$, then we must have $\mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt{-3})$.

Given that multiplicativity of degrees gives $\Big[ \mathbb{Q}(\sqrt{-3}): \mathbb{Q} \Big] = \Big[ \mathbb{Q}(\sqrt{-3}): \mathbb{Q}(\alpha) \Big] \cdot \Big[ \mathbb{Q}(\alpha): \mathbb{Q} \Big]$, think about the degrees of these extensions over $\mathbb{Q}$ to arrive at a contradiction.

Suppose it was reducible. Then it would have a root in $\mathbb Q(\sqrt{-3})$. But it is irreducible over $\Bbb Q$, so that would mean $\mathbb Q(\sqrt{-3})$ would contain an element of degree three over $\Bbb Q$. But it is an extension of degree two so that is impossible.

One way may be to using the counting theorem?

The degree of $\sqrt{-3}$ over $\mathbb{Q}$ is 2.

Since $x^3-3$ is of degree 3, if it was reducible in $\mathbb{Q}(\sqrt{-3})$, it would have a root $\alpha$ in $\mathbb{Q}(\sqrt{-3})$.

But by Eisenstein we also know that $x^3-3$ is irreducible over $\mathbb{Q}$. So assume for contradiction that it was reducible over $\mathbb{Q}(\sqrt{-3})$, and that $\alpha$ existed. Then we would have:

$[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}]=2=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)][\mathbb{Q}(a):\mathbb{Q}]=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)] \cdot 3$.

But $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)]$ must be an integer, so we have our contradiction from $2=[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)]\cdot3$.

In order to use this we also must have that $Q(\sqrt{-3})$ is a finite extension of $Q(\alpha)$. But since $Q(\sqrt{-3})$ is a finite extension of $\mathbb{Q}$, we have that $1, \sqrt{-3}$ is a basis for $Q(\sqrt{-3})$ over $\mathbb{Q}$. If $\sqrt{-3}$ should happen to be in $Q(\alpha)$ they must be the same, if not, we then have that $1, \sqrt{-3}$ over $Q(\alpha)$ span $Q(\sqrt{-3})$, but we also must have that $1, \sqrt{-3}$ must be linearly independent when having coefficients in $\mathbb{Q}(\alpha)$, if not $g_1+q_2\sqrt{-3}=0$, where not both coefficients in $Q(\alpha)$ is zero. But then it is easy to see that $\sqrt{-3}$ must be in $\mathbb{Q}(\alpha)$.

This last fact that $\mathbb{Q}(\sqrt{-3})$ is a finite extension of $\mathbb{Q}(\alpha)$ actually also follows from the fact that every simple extension when considering an element algebraic over another field is a finite extension, and $[\mathbb{Q}(\sqrt{-3}):\mathbb{Q}(\alpha)] $ is the degree of $(\mathbb{Q}(\alpha))(\sqrt{-3})$ over $\mathbb{Q}(\alpha)$. And we know that since $\sqrt{-3}$ is algebraic over $\mathbb{Q}$, it must also be algebraic over $\mathbb{Q}(\alpha)$.