induction - Show that $({sqrt{2}!+!1})^{1/n} !+ ({sqrt{2}!-!1})^{1/n}!notinmathbb Q$

How could we prove that for every positive integer $n$, the number
$$({\sqrt{2}+1})^{1/n} + ({\sqrt{2}-1})^{1/n}$$ is irrational?

I think it could be done inductively from a more general expression, but I don't know how.

I made an effort trying to solve it using many different methods.

Answer

Assume that for a certain positive integer $n$
$$\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\in \mathbb Q.$$
Clearly, $\,\big(\sqrt{2}-1\big)^{1/n}\cdot\big(\sqrt{2}+1\big)^{1/n}=1$, and thus
$$\left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right)^2= \big(\sqrt{2}-1\big)^{2/n}+2+\big(\sqrt{2}+1\big)^{2/n} \in \mathbb Q,$$
and hence
$$\big(\sqrt{2}-1\big)^{2/n}+\big(\sqrt{2}+1\big)^{2/n} \in \mathbb Q.$$
Next

$$\left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right)^3= \big(\sqrt{2}-1\big)^{3/n}+\big(\sqrt{2}+1\big)^{3/n}+3\left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right),$$
and hence
$$\big(\sqrt{2}-1\big)^{3/n}+\big(\sqrt{2}+1\big)^{3/n}\in\mathbb Q.$$
Let $s_k=\big(\sqrt{2}-1\big)^{k/n}+\big(\sqrt{2}+1\big)^{k/n}$. Then
$$s_1^k=s_k+\binom{k}{1}s_{k-2}+\binom{k}{2}s_{k-4} +\cdots+\binom{k}{\lfloor k/2\rfloor}s_{k-2\lfloor k/2\rfloor},$$
which means that we can inductively show that
$$s_1,s_2,\ldots,s_n\in\mathbb Q.$$
But
$$s_n=\big(\sqrt{2}-1\big)+\big(\sqrt{2}+1\big)=2\sqrt{2}\not\in\mathbb Q,$$
and hence
$$s_1=\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\not\in\mathbb Q.$$

Generalization. If $a,b\in\mathbb R$, such that
$a,b, a+b\not\in\mathbb Q$ and $ab\in\mathbb Q$, then using exactly this method, we obtain that
$a^{1/n}+b^{1/n}\not\in\mathbb Q$, for all $n\in\mathbb N$.