<s>[INST] Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational How can I prove that the sum  $\sqrt 2+\sqrt[3] 3$
 is an irrational number ??
 [/INST] If $x=\sqrt2+\sqrt[3]3$, then
$$
\begin{align}
3
&=(x-\sqrt2)^3\\
&=x^3-3\sqrt2x^2+6x-2\sqrt2\\
(x^3+6x-3)^2&=2(3x^2+2)^2\\
0&=x^6-6x^4-6x^3+12x^2-36x+1
\end{align}
$$
Thus, $x$ is an algebraic integer. Since $2\lt x\lt3$, $x\not\in\mathbb{Z}$, so $x\not\in\mathbb{Q}$. In this answer, it is shown that a rational algebraic integer is an integer.

Alternative approach
If $x=\sqrt2$ and $y=\sqrt[3]3$, then $x^2-2=0$ and $y^3-3=0$. Thus, both $x$ and $y$ are algebraic integers. Therefore, $x+y=\sqrt2+\sqrt[3]3$ is also an algebraic integer.
Note that this approach, while seeming simpler, requires that one knows that the sum of two algebraic integers is again an algebraic integer. This fact can be shown using some linear algebra.
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