$\sqrt{2} +\sqrt[3]{2}$ is irrational

You've got the right idea, but there are two (small) problems:

• the equation $$x^3-3x^2\sqrt{2}+6x+2\sqrt{2}=2$$ is wrong, the last sign on the left should be $-$, as in $(-1)^3=-1$.

• you should explain why it is possible to divide by $3x^2-2$. In fact, once you fix the first point I mentioned, it's going to be $3x^2+2$ and it's going to be much easier.

We can use field theory to conclude that the degree $$ [\Bbb Q(\sqrt{2}+\sqrt[3]{2}):\Bbb Q]=6>1. $$

The minimal polynomial of $\sqrt{2}+\sqrt[3]{2}$ is $x^6-6x^4-4x^3+12x^2-24x-4=0,$ see the first link.

References:

Finding the minimal polynomial of $\sqrt 2 + \sqrt[3] 2$ over $\mathbb Q$.

Finding a basis for the field extension $\mathbb{Q}(\sqrt{2}+\sqrt[3]{4})$

The first reference shows in a very elementary way that $\sqrt{2}\in \Bbb Q(\sqrt{2}+\sqrt[3]{2})$. Hence $\sqrt{2}+\sqrt[3]{2}$ is irrational.

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

$\tag 2 u^6 -6u^4-4u^3+12u^2 -24u - 4 = 0$

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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.