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 ??

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.
