Question: Let it be $\alpha=1+\sqrt[3]{5}\,i$ and $f(x)$ the minimal polynomial of $\alpha$ in $\mathbb{Q}$.

Is $\mathbb{Q}(\alpha)$ the splitting field of $\alpha$?

I generally know how to deal with splitting field when the roots are simple but I dont know how to deal with this or in general when minimal polynomial aren't that easy to handle with.

Are there any properties or theorems that help as criterions to determine whether we have a splitting field or not? (if there is anything more general that covers not only this case I would be more than pleased to note them down)

I hope I didn't spell the question wrong and thanks in advance

Answer: 
For your example, let $\beta=\sqrt[3]5i$, then we have $\beta^6=-25$. Clearly $\beta$ has degree 6 over $\mathbb Q$, and its conjugates are given by $\beta\zeta^k$, $k=0,\dots,5$, where $\zeta=\frac{1+\sqrt3i}2$ is a primitive sixth root of unity. Therefore, the extension $\mathbb Q(\beta)/\mathbb Q$ is not Galois, because $\zeta$ is not contained in $\mathbb Q(\beta)$.

If $\mathbb Q(\alpha)$ is the splitting field, then it must be a Galois extension, but $\mathbb Q(\alpha)=\mathbb Q(\beta)$. In general, the extension of the form $\mathbb Q(\sqrt[n]m)$ can be dealt with in the same way.
Knowing that ℚ(1+∛5i)=ℚ(∛5i) and that ∛5i is root of the polynomial x^6+25, then the splitting field must be ℚ(∛5i\*ω), where ω is a primitive root of x^6-1. Thus, ℚ(α) isn't the splitting field.