This is the orginal question 
option C will be true take $\mathbb{Q}[x] / (x^2 +1)$ that is subfield of $\mathbb{Q}[X]/(2X^3 − 3X^2 + 6)$
similarly option D) will aslo True same logics in option C
Im confused about option $A)$ and option $B)$
Any hints/solution will be apprecaited
thanks u
On
The cubic has one real and two complex roots (look at the discriminant).
(A) True: adjoin one of the complex roots (see my comment re typo)
(B) False: pick an irreducible cubic that has three real roots e.g. $x^3-4x+2$ (draw the graph to show it has three real roots, use Eisenstein to show it's irreducible)
(C) True: adjoin the real root to $\mathbb Q$
(D) True: adjoin the real root to $\mathbb Q$
Note that none of the extensions is normal.
Here are some hints that might help figure it out yourself:
i.) A cubic polynomial $f$ is irreducible over $\mathbb Q$ if and only if it has no roots in $\mathbb Q$
ii.) Making the polynomial monic doesn't change the field, i.e. $\mathbb Q[X] / (2X^3 - 3X^2 + 6) \cong \mathbb Q[X]/(X^3 - \frac{3}{2}X^2 + 3)$.
iii.) Note that if you have a monic irreducible polynomial $f$ and an element $\alpha \in \mathbb C$ with $f(\alpha) = 0$, then $f$ is the minimal polynomial of $\alpha$ and $\mathbb Q(\alpha) \cong \mathbb Q[X]/(f)$.
iv.) If you want to decide if such a field is contained in $\mathbb R$ or has elements that lie in $\mathbb C \setminus \mathbb R$, you will have to look at the roots. I think you can work out yourself how to decide that.
iv.) If you have roots $\alpha_1, \dots, \alpha_3$ of your polynomial $f$ above, note that you have $\mathbb Q(\alpha_1) \cong \mathbb Q(\alpha_2) \cong \mathbb Q(\alpha_3) \cong \mathbb Q[X]/(f)$. Hence you can look at all these three fields to decide if the statements above are true!
v.) Note that if a polynomial $f \in \mathbb Q[X]$ has a root in $\mathbb C \setminus \mathbb R$, then also the complex conjugate has to be a root. What does that tell you if $f$ is of degree 3?