<s>[INST] Can anyone explain why expressions of the form $\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}$ can be rational? $$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
Can anyone explain to me how this works? I don't understand why two irrational numbers cube rooted and added together return the number $1$.
I'm trying to find all the positive integer values for $x$ and $y$ in this formula
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
Sorry if this is a stupid question.
 [/INST] Note that
$$
\sqrt[3]{2+\sqrt{5}}=\frac{1}{2}\sqrt[3]{16+8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3+3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2+(\sqrt{5})^3}=\frac{1+\sqrt{5}}{2}.
$$
Similarly,
$$
\sqrt[3]{2-\sqrt{5}}=\frac{1}{2}\sqrt[3]{16-8\sqrt{5}}=\frac{1}{2}\sqrt[3]{1^3-3\cdot 1^2\cdot\sqrt{5}+3\cdot 1\cdot(\sqrt{5})^2-(\sqrt{5})^3}=\frac{1-\sqrt{5}}{2}.
$$
Thus,
$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1,
$$
as desired.
As for the second question, we can use the @player3236's approach. Let
$$
a=\sqrt[3]{x+\sqrt{y}},~b=\sqrt[3]{x-\sqrt{y}}.
$$
Then, for $c=a+b$ we have
$$
c^3=2x+3c\sqrt[3]{x^2-y},
$$
so $c$ is a root of the polynomial $t^3-3\sqrt[3]{x^2-y}\cdot t-2x$. If $c=a+b=1$, then
$$
1-3\sqrt[3]{x^2-y}-2x=0,
$$
or
$$
\sqrt[3]{y-x^2}=\frac{2x-1}{3},
$$
or
$$
y=\frac{1}{27}(27x^2+(2x-1)^3).
$$
We've obtained some necessary condition. However, I'm not sure whether this condition is sufficient or not (I guess that proof should be similar to case $x=2$, $y=5$, which was discussed above).
</s>