Let $f(x)$ be a polynomial of degree 3 over rationals. What are the possible values of [$\mathbb{Q}(a_1,a_2,a_3)$:$\mathbb{Q}$], where $a_i$ are roots of $f(x)$.
I was managed to get 1,2,6 but i don't know exactly whether this value could be equal to 3,4,5
We know that the degree of the splitting field of some polynomial of degree $n$ divides $n!$. In our case $n=3$, so $[\mathbb{Q}_{f}:\mathbb{Q}]|6$. So the possible values are $1,2,3,6$.
The polynomial has all of his roots in the field of the rationals, for example $f=(t-3)^2$
For example something like $f=(t-1)(t^2+1)$, in that case $\mathbb{Q}_f=\mathbb{Q}(i)$, or for example $f=t^3-1$. In that case $\mathbb{Q}_f=\mathbb{Q}(\zeta)$, where $\zeta=e^{2\pi i/3}$.
For example $f=8^3-6t+1$. In that case $\mathbb{Q}_f=\mathbb{Q}(a)$, where $a$ is some root of the polynomial. The interesting thing about that polynomial is that the other roots are expressed in terms of $a$, you can prove easily that the roots are: $a$, $2a^2-1$ and $-2a^2-a+1$
For example $f=t^3-2$, in that case $\mathbb{Q}(\zeta,\sqrt[3]{2})$, where $\zeta$ is as defined previously.
To prove the interesting thing, that the degree of the extension divides $n!$ let's prove it by induction over the number of roots $\alpha_1,\cdots,\alpha_n$:
Consider the extension $\mathbb{Q}(\alpha_1)$. Since $\alpha_1$ is a root of your polynomial $f(t)$, we know that $f(t)=(t-\alpha_1)g(t)$ (because $f(t)$ factors in $\mathbb{Q}(\alpha_1)[x]$) where $g(x) \in \mathbb{Q}(\alpha_1)[t]$. Hence the minimal polynomial of $\alpha_2$ over $\mathbb{Q}(\alpha_1)$ must divide $g(x)$ (which has degree $n-1$), so $[\mathbb{Q}(\alpha_2,\alpha_1):\mathbb{Q}(\alpha_1)] \leq n-1$. The result follows by induction.