Proving $\sqrt{2}+\sqrt[3]{3}$ is algebraic.

Question

How can I prove that $\sqrt{2}+\sqrt[3]{3}$ is algebraic?

I have been trying various things, and have failed as of now. I know that the degree of the polynomial satisfying this equation has to be $\leq 6$. But that is all I know. Clearly, the simple idea of building the polynomial from the root itself, like in the case of $\sqrt{2}+\sqrt{3}$ does not work here.

Thanks in advance!

On the contrary, building a polynomial from the root works perfectly! — Gerry Myerson, Feb 13 '14 at 08:39
It may be more instructive to prove in general that the sum of algebraics is algebraic — Hagen von Eitzen, Feb 13 '14 at 08:40
it is enough to show that is is algebraic over $\mathbb{Q}(\sqrt{2})$... — Blah, Feb 13 '14 at 08:59

score 8 · Answer 1 · answered Feb 13 '14 at 08:44

Let

$$x=\sqrt 2+\sqrt[3] 3$$ then $$(x-\sqrt 2)^3=x^3-3x^2\sqrt2+6x-2\sqrt 2=3\iff x^3+6x-3=\sqrt2(3x^2+2)$$ now square the two sides $$(x^3+6x-3)^2=2(3x^2+2)^2$$ and simplify you find that $x$ is a root of a polynomial with integer coefficients.

score 2 · Answer 2 · answered Feb 13 '14 at 08:43

2

If $x=\sqrt 2+\sqrt[3]3$ then $(x-\sqrt 2)^3=3$, i.e. $$ x^3-3\sqrt 2 x^2+6x-2\sqrt 2=3$$ or $$ x^3+6x -3=(3x^2+2)\sqrt 2.$$ Now square

answered Feb 13 '14 at 08:43

Hagen von Eitzen

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Proving $\sqrt{2}+\sqrt[3]{3}$ is algebraic.

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