# 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 $$x^3-18x-110$$.

• Could you show in detail your attempt using conjugate pairs? – TheSimpliFire Oct 11 '19 at 15:49
• Yeah, like I tried starting with roots as (x-2^1/3-3*4^1/3)(x-2^1/3+3*4^1/3) – Toshu Oct 11 '19 at 15:52
• Great, please put that in your post to prevent it from being closed. Try to include as much detail as possible. – TheSimpliFire Oct 11 '19 at 15:53
• Yes I'm putting it my post, just a second – Toshu Oct 11 '19 at 16:00

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

• If ∆ is a root, then to make a polynomial can I take x=∆ like you took x = y+3y^2? – Toshu Oct 11 '19 at 16:19
• Yes, because $(x-\Delta)$ would be a factor of the polynomial. – TheSimpliFire Oct 11 '19 at 16:21
• You mean like as x-∆=0 (as it is a root) x= ∆ right – Toshu Oct 11 '19 at 17:06
• You are correct. – TheSimpliFire Oct 11 '19 at 17:06

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

$$[\Bbb Q(\alpha):\Bbb Q] < 3, \tag{19}$$

• Great Answer 👍🏻. But can you please tell me what you are proving after (10) equation? – Toshu Oct 12 '19 at 2:30
• It seems that TheSimpliFire also added something like this to his answer. – Toshu Oct 12 '19 at 2:31
• @UnnayanUpadhyay: Sorry it took me so long to get back to you. Up until (9), we have shown that $\alpha$ satisfies the cubic polynomial $m_\alpha(x) = x^3 - 18x - 110 \in \Bbb Q[x]$, but still haven't addressed the minimality of $m_\alpha(x)$; this is the subject of my answer from (10) on; I do this in essence by showing that $[\Bbb Q(\alpha):\Bbb Q] = [\Bbb Q(\sqrt[3]2)] = 3$, so $\alpha$ cannot satisfy a polynomial of degree $2$ since $\not \mid 3$; and neither of degree $1$, as is explained in my answer. Hope this helps. Feel free to ask more questions. Cheers! – Robert Lewis Oct 13 '19 at 1:26