Show $(3 + \sqrt{2})^{2/3}$ is irrational using RZT

Question

I am asked to prove that $(3+\sqrt{2})^{2/3}$ is irrational via the rational zeroes theorem.

This is what I have so far:

$ x = (3+\sqrt{2})^{2/3} $

$ x^3 = (3+\sqrt{2})^{2} $

$ x^3 - 11 - 6\sqrt{2} = 0 $

From here, I do not know how to get it into the form admissible by the RZT. The only way I know how to proceed from here is to say that any rational solution of the form $r= \frac{c}{d}$ $c,d, \in \mathbb{Z} $ would need to have $d= \pm 1$ and $c$ divide $-11 -6\sqrt{2}$.

So we have

$ -11 -6\sqrt{2} = z c$, $z \in \mathbb{Z}$

$-6\sqrt{2} = 11 +zc$

Which clearly no $c$ will satisfy. So it is proved, but I feel like I did not utilize the RZT the way I was supposed to. At least, it is different than the other example problems I have been doing, for example, proving $\sqrt{3}$ is irrational.

Answer 1

From $x^3-11-6\sqrt2=0$, you get that $(x^3-11)^2=72$; in other words, $x^6-22x^3+49=0$. But the only possyble rational roots of this polynomial are $\pm1$, $\pm7$, and $\pm49$. However, none of them is.

Answer 2

As an alternative, we have that

$$x=(3 + \sqrt{2})^2=13+6\sqrt 2\not \in \mathbb{Q}$$

now suppose by contradiction that

$$\sqrt[3] x =y\in \mathbb{Q} \iff y^3=x$$

which is impossible.

Answer 3

You don't need the RZT and you already did all the hard work: from

$$x^3-11-6\sqrt2=0$$

we can deduce as follows: if $\;x\in\Bbb Q\;$ then also $\;x^3\in\Bbb Q\;$ and thus also $\;x^3-11=6\sqrt2\in\Bbb Q\;$ , from where we get the straightforward contradiction that $\;\sqrt2\in\Bbb Q\;$ ...

Answer 4

Note $(3 + \sqrt 2)^{\frac 23}$ doesn't have to be the only root.

You got that it is a solution to

$x^3 - (11 + 6\sqrt 2) = 0$.

So it is a solution to $(x^3 - (11 + 6\sqrt 2))(x^3 - (11 - 6\sqrt 2)) =$

$=x^6 - 22x^3 - (11^2 - 72) = x^6 -22x^3 - 49=0$.

Now by the rational root test the only possible rational roots are $\pm 1, \pm 7, \pm 49$ and not of them are roots.

====

D'oh. Just read Jose Carlos Santos answer. Yeah $x^3 - 11 -6\sqrt2=0 \iff (x^3 -11) = 6\sqrt 2 \implies (x^2 - 11)^2 = 72$ is a lot more obvious an insightful and easier than my idea of looking for conjugates.

They both work and have the same result but in terms of ease in seeing and teaching... His is better.

(Although theoretically his isolating a square root and squaring it and my multiplying by conjugates is basically the same thing.)