Process to show that $\sqrt 2+\sqrt[3] 3$ is irrational

If $x=\sqrt{2}+\sqrt[3]3$, then

$$(x-\sqrt{2})^3=x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=3$$

Thus

$$x^3+6x-3=\sqrt{2}(3x^2+2)$$

And

$$\frac{x^3+6x-3}{3x^2+2}=\sqrt{2}$$

But if $x$ is a rational, then so is the left hand side of the above equality. However we know $\sqrt{2}$ is not rational. Contradiction, so $x$ is irrational.

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.

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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.

The following is a solution using some (basic) notions from the theory of field extensions:

Assume $\sqrt{2}+\sqrt[3]{3}$ is rational. Then $\sqrt[3]{3}\in\mathbb{Q}(\sqrt{2})$, thus $\sqrt[3]{3}$ is a root of some quadratic polynomial over $\mathbb{Q}$ which leads to a contradiction, since the polynomial $x^3-3$ is irreducible over $\mathbb{Q}$.

The same idea can be expressed using only elementary terms. Let $a$ be a rational number, and note that $a-\sqrt{2}$ is a root of the rational polynomial $x^2-2ax+a^2-2$. Thus, if $\sqrt{2}+\sqrt[3]{3}$ is rational, it follows that $\sqrt[3]{3}$ is a root of a quadratic rational polynomial, which is a contradiction as $x^3-3$ is irreducible.