Proving $1 +\sqrt[3]{5-\sqrt2}$ is rational via the rational roots theorem

Question

Find a polynomial that has $x = 1 +\sqrt[3]{5-\sqrt2}$ as a root, then use rational roots theorem to show that $x$ must be rational.

I came up with the polynomial $100,000,000,000x^2 = 640393215809$ but their GCD is only $1$, so it won't work for part 2. Anyone help with an easier function.

Answer 1

How about $$x=1+\sqrt[3]{5-\sqrt 2}\\ x-1=\sqrt[3]{5-\sqrt 2}\\ (x-1)^3=5-\sqrt 2\\ \sqrt 2=5-(x-1)^3\\ 2=\left(5-(x-1)^3\right)^2\\ 0 = x^6 - 6 x^5 + 15 x^4 - 30 x^3 + 45 x^2 - 36 x + 34$$ but I don't think your number is rational. Alpha does not find it to be. I suspect the problem is to prove your number is irrational by the rational root theorem by finding all the possible rational roots of this equation and noting that your number is none of them.

Answer 2

If $1 +\sqrt[3]{5-\sqrt2}$ were rational, then $\sqrt[3]{5-\sqrt2}$ would be rational, so its cube $5-\sqrt2$ would be rational. Which it isn't.

Answer 3

I got the minimal polynomial of your number to be $$ 34 - 36X + 45X^2 - 30X^3 + 15X^4 - 6X^5 + X^6\,, $$ which is not zero at $\pm1,\pm2\pm17$ or $\pm34$, so your number is irrational.

Answer 4

$((x\!-\!1)^{\large 3}\!-\!5)^{\large 2} = \color{#c00}2\,$ so if $\,x\in\Bbb Q,\,$ RRT $\,\Rightarrow\,x\in \Bbb Z\,\Rightarrow\,\sqrt{\color{#c00}2}\in\Bbb Z,\,$ contradiction, so $\,x\not\in\Bbb Q$