Simplifying equation with different radicals

Question

I am trying to simplify the following expression, so it can be written only with rational coefficients:

$$x = \sqrt{2} + \sqrt[3]{5}$$

I've tried multiple basic operations, such as moving the radicals to one side and powering it to 2 or 3, but as I develop the expression I only get bigger and bigger radicals until I reach the point that I can't handle them.

What would be the winning approach here? Thanks.

For approaches see some posts here, e.g., this one, or this one. — Dietrich Burde, Nov 18 '21 at 10:53
"so it can be written only with rational coefficients" ==> What do you mean by that ? What do you mean by the word "coefficient" ? — G. Fougeron, Nov 18 '21 at 10:59
By coefficient I meant the term that multiplies x, x^2, etc.
Thanks! I actually found something similar to what I was looking for in @Dietrich Burde linked post. — thegreenhoodie, Nov 18 '21 at 11:06
Since you have a square root and a cube root, you should expect that you would need a degree 6 polynomial -- and indeed this is what Galois theory tells you. — Kenny Lau, Nov 18 '21 at 11:13

score 2 · Accepted Answer · answered Nov 18 '21 at 12:52

You won't be able to "simplify your expression", as $x$ is not rational. However, if you want to identify $x$ as a root of polynomial $f$ with rational coefficients (that is $f(x)=0$) , you are on the right track.

Start with $x-\sqrt{2}=\sqrt[3]{5}$, then you cube everything to get an expression of the form $a+b\sqrt{2}=5$ where $a,b$ polynomial expressions in $x$ with rational coefficients. Then, $a-5=-b\sqrt{2}$ , and thus $(a-5)^2=2b^2$.

All in all, you will find that $x^6 - 6 x^4 - 10 x^3 + 12 x^2 - 60 x + 17=0$.

Simplifying equation with different radicals

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