$\sqrt[3]{2}$ satisfies $x^3-2=0$ Show that there is no polynomial $P(x)$ of degree less than $3$ with $P(\sqrt[3]{2})=0$ All coefficients are rational numbers.

Is it by induction? Say, if $x$ has degree of $1$, it doesn't work; and for $x$ having degree of $2$, am I applying Polynomial remainder theorem?

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    $\begingroup$ Welcome to Maths SX! Do you mean $3\sqrt 2$ or $\sqrt[3]2$? $\endgroup$
    – Bernard
    Oct 7 '19 at 19:24
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    $\begingroup$ You mean $\sqrt[3]{2}$. And this is the minimal polynomial, see the duplicates. $\endgroup$ Oct 7 '19 at 19:25
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    $\begingroup$ As stated, it’s false whether you meant $3\sqrt2$ or $\sqrt[3]2$, since whatever value $a$ you have, it’s a root of $P(x)=x-a$. Perhaps there’s some restriction on the coefficients of the polynomials that you haven’t mentioned? $\endgroup$
    – amd
    Oct 7 '19 at 19:28
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    $\begingroup$ You need to specify the field your polynomial is over. For example the polynomial $x-2^{1/3}$ would be a counterexample to your claim if the field was arbitrary. I assume the field is $\Bbb{Q}$. As for your task, just factor $(x-2^{1/3})$ out from your polynomial and show that the some coefficient of the other factor is not in $\Bbb{Q}$. $\endgroup$
    – B.Swan
    Oct 7 '19 at 19:32
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    $\begingroup$ Look up minimal polynomial. The polynomial $x^3-2$ is irreducible over $\Bbb{Q}$, so it must be the minimal polynomial of $\root3\of2$, i.e. the lowest degree polynomial with rational coefficients vanishing at $\root3\of2$. By the way, you need to specify the field of coefficients. See amd's comment as well as that of B.Swan. $\endgroup$ Oct 7 '19 at 19:33

You didn't precised what are the nature of the coefficients of P, so we can build a polynomial $P(x) = x - 3\sqrt{2}$ such that $P(3\sqrt{2}) = 0$.

And, we can build an infinity of polynomials of degree 2 such that $P(2 \sqrt{3}) = 0$, these polynomials are : $$P(x) = (x - 2 \sqrt{3})(x - k), k \in \mathbb{C}$$

We all must be very careful to mathematical rigor.

If we consider that $P \in \mathbb{Z}[X]$, it is obvious that there is no polynomials of degree 1 such that $P(2 \sqrt{3}) = 0$ because $\forall P \in \mathbb{Q}_1[X], P(X) = 0 \implies X \in \mathbb{Q}$, and, because $\mathbb{Z}_1[X] \subset \mathbb{Q}_1[X]$, this property is true even for polynomials with relative coefficients (and $2 \sqrt{3} \notin \mathbb{Q}$).

Then, for polynomials with degree 2, we use the quadratic formula :

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$, so we search solution to the equation :

$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = 2 \sqrt{3}, (a,b,c) \in \mathbb{Z}^3$$

We can directly see that we must have $b = 0$ because if that was not the case, we should have $x = k + l \sqrt{m}, k \neq 0$ knowing that b must be a relative number. Or, we see that the solution we are searching for is not looking like that.

So, because we have $b = 0$, we have $$x = \frac{\pm \sqrt{-4ac}}{2a}$$ with one of these two solutions equal to $2 \sqrt{3}$. Because we have $2\sqrt{3} \notin \mathbb{C} - \mathbb{R}$, we must have $-4ac > 0$ so $4ac < 0$.

We have so $$x = \pm 2 \frac{\sqrt{-ac}}{2a} = \pm \frac{\sqrt{-ac}}{a}$$ Then, we suppose $a > 0$ so we can write $$x = \pm \frac{ \sqrt{a} }{a} \times \sqrt{-c} = \pm \frac{\sqrt{-c}}{\sqrt{a}}$$ so we must have $\sqrt{-c} = \sqrt{3}$ and $1/\sqrt{a} = \pm 2$ which is not possible with $a \in \mathbb{Z}$

And, if we have $a < 0$, we have $$x = \pm \frac{\sqrt{c}}{\sqrt{-a}}$$

So we must have $c = 3$ and $1/\sqrt{-a} = \pm 2$ which is not possible with $a \in \mathbb{Z}$, QED.

EDIT : The proof is also valable for $x = 2^{1/3}$ and $3\sqrt{2}$


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