Exercise 1(d) from Courant I'm having trouble understanding this "hint" in the back of (the first volume of) Courant's Differential and Integral Calculus text, which I'm just starting:  
One of the "challenging" Chapter 1 exercises asks you to prove that $x = \sqrt{2} + \sqrt[3]{2}$ is irrational; the hint says to "Show that $x$ satisfies an equation of the type
$$
x^6 + a_1x^5 + ...+ a_6 = 0
$$
where $a_1, ..., a_6$ are integers; prove that $x$ is then either irrational or an integer. 
I guess I don't really understand what one is supposed to be "showing" here - and even if one does in fact "show" what Courant is expecting you to "show", musn't $x$ necessarily be rational according to Gauss' lemma?
 A: Hint: Make the polynomial, perhaps by starting from $x-\sqrt{2}=\sqrt[3]{2}$ and cubing both sides.  
By the Rational Roots Theorem any rational root will have to be of the shape $\frac{c}{d}$, where $c$ is a divisor of $a_6$ and $d$ is a divisor of $a_0$, in this case $1$. There is only a finite number of candidates. If none works, you will know that the sixth degree polynomial has no rational roots.
Remark: You asked about one particular approach to the problem. A slightly different way is to let $x=\sqrt{2}+\sqrt[3]{2}$. 
We have $x-\sqrt{2}=\sqrt[3]{2}$. Cube both sides. We get $x^3-3\sqrt{2}x^2+6x-2\sqrt{2}=2$.
Solving for $\sqrt{2}$ we find that 
$$\sqrt{2}=\frac{x^3+6x-2}{3x^2+2}.$$
If $x$ was rational, then, from the above expression. $\sqrt{2}$ would be rational. But it isn't. So $x$ cannot be rational.
However, what you are intended to do is to write $x^3+6x-2=(3x^2+2)\sqrt{2}$, square both sides, and simplify. You then get the degree $6$ equation with integer coefficients mentioned in the book. 
A: By Gauss Lemma, if the root is a rational, it must be an integer, and you can show that your $x$ is not an integer. And hence, it cannot be rational.
