Find the polynomial equation of the lowest degree with rational coefficients whose one root is.......? Find the polynomial equation of the lowest degree with rational coefficients whose one root is $\sqrt[3]{2}+3\sqrt[3]{4}$
I tried using the conjugate pairs but I couldn't solve it for any polynomial equation other than one having roots to the power 1/2.
I took the roots as $(x-\sqrt[3]{2}-3\sqrt[3]{4})(x-\sqrt[3]{2}+3\sqrt[3]{4})$ but after a few multiplications (taking conjugates of the polynomial repeatedly) the roots would become too complicated and the degree would rise to be more than 6.
A detailed Explanation would be helpful.
The Answer is $x^3-18x-110$.
 A: Let $y=\sqrt[3]2$. Then $x=y+3y^2=y(3y+1)$ so cubing both sides yields $$x^3=y^3(27y^3+27y^2+9y+1)=2(27\cdot2+9y(3y+1)+1)=2(9x+55)$$ so $x^3-18x-110=0$. This is the minimal polynomial as $[\Bbb Q(\sqrt[3]2):\Bbb Q]=3$.
A: Lauds to our colleague TheSimpliFire for a concise and elegant answer, most worthy of the coveted green check of acceptance.
In contrast, I present a klunky and grungy elementary though computationally laborious approach:
Let
$\alpha = \sqrt[3]2 + 3\sqrt[3]4; \tag 1$
then we may compute $\alpha^3$ via an elemntary but somewhat tedious calculation which is highly remeiscient of high-school algebra; first, by the binomial theorem applied to $(\sqrt[3]2 + 3\sqrt[3]4)^3$:
$\alpha^3 = 2 + 3(\sqrt[3]2)^2(3\sqrt[3]4) + 3(\sqrt[3]2)(3\sqrt[3]4)^2 + (27)(4); \tag 2$
next it's just a simple matter of simple arithmetic and reducing and resolving the radicals:
$\alpha^3 = 2 + 9(\sqrt[3]4)^2 + 27(\sqrt[3]2)(\sqrt[3]{16}) + 108; \tag 3$
$\alpha^3 = 2 + 9\sqrt[3]{16} + 27\sqrt[3]{32} + 108; \tag 4$
$\alpha^3 = 110 + 9\sqrt[3]{8 \cdot 2} + 27\sqrt[3]{8 \cdot 4}; \tag 5$
$\alpha^3 = 110 + 18\sqrt[3]2 + 54\sqrt[3]4; \tag 6$
$\alpha^3 = 18(\sqrt[3]2 + 3\sqrt[3]4) + 110; \tag 7$
we substitute (1) on the right;
$\alpha^3 = 18\alpha + 110, \tag 8$
or
$\alpha^3 - 18\alpha - 110 = 0; \tag 9$
note that
$\alpha \in \Bbb Q(\sqrt[3]2) \tag{10}$
and that
$[\Bbb Q(\sqrt[3]2):\Bbb Q] = 3; \tag{11}$
since
$[\Bbb Q(\sqrt[3]2:\Bbb Q(\alpha)][\Bbb Q(\alpha):\Bbb Q] = [\Bbb Q(\sqrt[3]2):\Bbb Q] = 3, \tag{12}$
it follows that
$[\Bbb Q(\alpha):\Bbb Q] = 1 \; \text{or} \; 3; \tag{13}$
since
$\alpha \notin \Bbb Q, \tag{14}$
we rule out
$[\Bbb Q(\alpha):\Bbb Q] = 1, \tag{15}$
and so
$[\Bbb Q(\alpha):\Bbb Q] = 3; \tag{16}$
whence
$[\Bbb Q(\sqrt[3]2:\Bbb Q(\alpha)] = 1 \Longrightarrow \Bbb Q(\sqrt[3]2) = \Bbb Q(\alpha); \tag{17}$
we then see that the polynomial
$m_\alpha(x) = x^3 - 18x - 110 \in \Bbb Q[x] \tag{18}$
must indeed be minimal for $\alpha$ over $\Bbb Q$, since $\alpha$ may satisfy no polynomial in $\Bbb Q[x]$ of degree less than $3$, lest 
$[\Bbb Q(\alpha):\Bbb Q] < 3, \tag{19}$
in contradiction to (16).
