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How to show that $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ is algebraic?

I know how to prove x = $\sqrt[3]{3} + \sqrt{2}$ and x = $\sqrt{3} + \sqrt{2}$ , but the root doesn't disappear from $\sqrt[3]{3} + \sqrt{3} + \sqrt{2}$ .

I want to make an algebraic equation that has this as an solution.

yunyoung
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    I don't suppose you can use the fact that the algebraic numbers form a field? Do you need to find a polynomial with that as a root? Also: $\sqrt[3]{3}$ gives $\sqrt[3]{3}$. – Brian Tung Oct 05 '22 at 07:55
  • It's not clear what the context of the problem is (e.g., whether you have any tools of field theory available), so I won't mark as a duplicate, but cf. https://math.stackexchange.com/questions/141427/sums-and-products-of-algebraic-numbers, https://math.stackexchange.com/questions/155122/how-to-prove-that-the-sum-and-product-of-two-algebraic-numbers-is-algebraic – Travis Willse Oct 05 '22 at 07:56
  • $\sqrt[3]{3}$ is integral over $\Bbb Z$, and also $\sqrt{2}$ and $\sqrt{3}$. Hence their sum is integral, since integral elements over $\Bbb Z$ form a ring. So it is algebraic. – Dietrich Burde Oct 05 '22 at 07:57
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    Does this answer your question? How to prove $\sqrt{2} + \sqrt[3]{2}$ is an algebraic number – Dietrich Burde Oct 05 '22 at 08:04
  • Thank you for your help. I tried that method, but I failed... – yunyoung Oct 05 '22 at 08:08
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    I am guessing that the OP (i.e. original poster) is attempting to derive a polynomial $$f(x) = a_nx^n + a_{n-1}x^{(n-1)} + \cdots + a_1x^1 + a_0,$$ such that $$a_n, a_{n-1}, \cdots, a_1,a_0$$ are all integers, such that they are not all $(0)$, and such that $$f\left[\sqrt[3]{3} + \sqrt{3} + \sqrt{2}\right] = 0. $$ If my guess is accurate, then my first try would be to set $$n=6,$$ chart out the values $$\left[\sqrt[3]{3} + \sqrt{3} + \sqrt{2}\right]^k ~: ~k \in {1,2,3,4,5,6},$$ and then try to use the chart to derive the satisfying integers $$a_6, a_5, \cdots, a_1, a_0.$$ – user2661923 Oct 05 '22 at 08:19
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    Using the approach in this answer, we can show that $\sqrt[3]3+\sqrt3+\sqrt2$ is a root of $x^{12}-30 x^{10}-12 x^9+303 x^8-1006 x^6-1656 x^5+1113 x^4+5172 x^3+2940 x^2+144 x-8648=0$ – robjohn Oct 05 '22 at 09:43

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You can show that the polynomial of degree $2\cdot 2 \cdot 3 = 12$

$$\prod_{\epsilon^2 = \epsilon'^2 = \omega^3=1} ( x -\epsilon \sqrt{2} -\epsilon' \sqrt{3} - \omega \sqrt[3]{3})$$ has integral coefficients and your number as a root.

In general you cook up a polynomial formed from all possible sums of roots in a similar way ( here use roots of $x^2-2$, $x^2-3$, $x^3-3$ ).

More hints: the above polynomial equals

$$\prod ((x\pm \sqrt{2}\pm \sqrt{3})^3 - 3)$$

orangeskid
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  • Oh, I think I'm misunderstanding something. I think it's an equation that has that as the solution, but can I make all the coefficients of the equation rational? – yunyoung Oct 05 '22 at 08:54
  • @yunyoung: Expand the brackets and get integral (so rational) coefficients – orangeskid Oct 05 '22 at 08:56
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Here is a Simpler Approach , which intuitively shows that we will have Integer Co-efficient Equation.

Let $(x-\sqrt{2}+\sqrt{3})=(\sqrt[3]{3})$

Cubing both sides , we get :
Let $(x-\sqrt{2}+\sqrt{3})^3=(3)$

Here I will use $@$ to indicate some arbitrary Co-efficient having only Integers or $x$ or Powers. Every $@$ is a new instance , not connected to every other $@$.

In case I have to refer to the actual value , I will use $A,B,C,D$ to indicate some arbitrary Co-efficient having only Integers or $x$ or Powers.

When we expand the Cube, we will get :
$(@+@\sqrt{2}+@\sqrt{3}+@\sqrt{6})=(3)$

It looks like we got rid of $\sqrt[3]{3}$ , but introduced $\sqrt{6}$ !

Move the terms around , to get this :
$(A+B\sqrt{6})=(@\sqrt{2}+@\sqrt{3})$

Multiply both sides by $(A-B\sqrt{6})$
We get :
$(A^2-6B^2)=(@\sqrt{2}+@\sqrt{3})$ & we have got rid of the $\sqrt{6}$ !

Square both sides , to get :
$(@)=(@+@\sqrt{6})$ & we have again introduced $\sqrt{6}$ while getting rid of $\sqrt{2}$ & $\sqrt{3}$ !

Move the terms around to get this :
$(C)=(D\sqrt{6})$

Squaring both sides again , we get the Integer Co-efficient Equation which has the given root & other roots :
$(C^2)=(6D^2)$

The Degree is $3 \times 2 \times 2 \times 2 = 24$

Hence the given number is Algebraic.

Prem
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