<s>[INST] $\sqrt{2} +\sqrt[3]{2}$ is irrational 
Prove that $\sqrt{2} +\sqrt[3]{2}$ is irrational.

My attempt: Suppose $\sqrt{2} +\sqrt[3]{2}$  is rational then $\exists$ $x\in \mathbb{Q}$ such that $$\sqrt{2} +\sqrt[3]{2}=x$$ Rewriting the above equation as $$x-\sqrt{2}=\sqrt[3]{2}$$ cubing the above equation gives us $$x^3-3x^2\sqrt{2}+6x-2\sqrt{2}=2$$ This implies that $$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}\in\mathbb{Q}$$ but this is absurd . Therefore ,$\sqrt{2} +\sqrt[3]{2}$  is irrational. Does this look good ?  Can this be done any other way.
This question is from chapter 2 of Spivak's Calculus.  In there he's given a hint which says "start by working with the 6th power " of this expression.. I don't see how that helps . Can you show me how it can be done using the hint the Author has provided ? Thank you
 [/INST] The OP shows great ingenuity with their (now fixed up) answer.

You can also go with Spivak's hint:

From Spivak's Exercise 18.a we know that if $u$ satisfies
$\tag 1 u^n + a_{n-1}u^{n-1} + \dots + a_0 = 0$
for integers $a_{n-1}, \dots, a_0$, then $u$ is either irrational or an integer.
Let
$\quad u = 2^{\frac{2}{6}} + 2^{\frac{3}{6}}$
It can be show that $u$ isn't an integer (see next section).
Moreover, it can be shown (by solving a system of linear equations) that
$\tag 2 u^6 -6u^4-4u^3+12u^2 -24u - 4 = 0$

Since
$\quad \sqrt{2} \lt \frac{3}{2}$
$\quad \sqrt[3]{2} \lt \frac{4}{3}$
we can write
$\quad \sqrt{2} + \sqrt[3]{2} \lt \frac{3}{2} + \frac{4}{3} \lt 3$
Since
$\quad \sqrt{2} \gt \frac{5}{4}$
$\quad \sqrt[3]{2} \gt \frac{5}{4}$
we can write
$\quad \sqrt{2} + \sqrt[3]{2} \gt \frac{5}{4} + \frac{5}{4} \gt 2$
So $\sqrt{2} + \sqrt[3]{2}$ can't be an integer.
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