Proving that $\sqrt[3] 7 -\sqrt 2$ is irrational. I understand proving that $\sqrt{7}-\sqrt {2}$ is irrational, but how does the answer change if its cube root of $7$ instead of square root?
the way I solve $\sqrt{7}-\sqrt {2}$ is by assuming its rational, and multiplying both sides by $\sqrt{7}+\sqrt {2}$, and deriving a contradiction
Thank you!
 A: $\mathbb{Q}(\sqrt[3]{7})$ is a degree 3 extension of $\mathbb{Q}$.
Suppose that $\sqrt[3]{7} - \sqrt{2}$ is rational (call it $a$). Then, $\sqrt[3]{7} = \sqrt{2} + a$. Therefore,  $\mathbb{Q}(\sqrt[3]{7}) = \mathbb{Q}(\sqrt{2})$.
However, $\mathbb{Q}(\sqrt{2})$ is a degree 2 extension of $\mathbb{Q}$. This is a contradiction, as the same extension cannot be both degree 2 and degree 3.
A: Let $\alpha=\sqrt[3] 7 -\sqrt 2$.
Then $\alpha$ is a root of $x^6 - 6 x^4 - 14 x^3 + 12 x^2 - 84 x + 41=0$.
Since this is a monic polynomial with integer coefficients (the actual polynomial is not important), the rational root theorem tells you that $\alpha$ is either irrational or an integer.
Now
$\quad 1.9 <\sqrt[3] 7 < 2.0 $
$\quad 1.4 < \sqrt 2 < 1.5 $
gives
$\quad 0.4 < \alpha < 0.6 $
which proves that $\alpha$ is not an integer and so must be irrational.
We can avoid these fine estimates: if $\alpha$ is an integer then it must divide $41$ (now the polynomial is important), but clearly $0 < \alpha < \sqrt[3] 7 < 2 $ and $\alpha \ne 1$.
A: Let’s go for an elementary proof (no rational root theorem and no field extension)
Assume $r=\sqrt[3]{7}-\sqrt{2}$ is rational. This means $\sqrt[3]{7}=r+\sqrt{2}$. Cubing one gets
$$7=r^3+3r^2\sqrt{2}+6r+2\sqrt{2}$$
This leads to
$$\sqrt{2}={7-r^3-6r\over 3r^2+2}$$
$r$ being rational this means $\sqrt{2}$ is rational. A contradiction
