Find the polynomial equation of the lowest degree with rational coefficients whose one root is.......?

Question

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 $x^3-18x-110$.

Answer 1

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 $[\Bbb Q(\sqrt[3]2):\Bbb Q]=3$.

Answer 2

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).