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.
 A: Writing $c = \sqrt[3]{2-\sqrt{5}}, d = \sqrt[3]{2+\sqrt{5}}, a=c+d$, cubing the expression gives:
\begin{align}(c+d)^3 &= c^3+3c^2d+3cd^2+d^3\\
&=(2-\sqrt 5) + 3cd(c+d) + (2+ \sqrt 5)\\
&=4+3a\sqrt[3]{(2-\sqrt5)(2+\sqrt 5)}\\
&=4 -3a=a^3
\end{align}
Hence $a^3+3a-4 = 0 = (a-1)(a^2+a+4)$.
The quadratic equation $a^2+a+4=0$ has no solutions in the reals. Hence $a=1$.
Hopefully this answer can give some insight on how you can approach the general problem for $\sqrt[3]{x\pm\sqrt y}$.

Update: Since the other answer has proceeded to partially solve your second question, I may as well prove the sufficient condition.
Writing $a = \sqrt[3]{x-\sqrt y}, b =\sqrt[3]{x+\sqrt y}, c=a+b$:
\begin{align}(a+b)^3 &= a^3+3a^2b+3ab^2+b^3\\
&=(x-\sqrt y) + 3ab(a+b) + (x+ \sqrt y)\\
&=2x+3c\sqrt[3]{(x-\sqrt y)(x+\sqrt y)}\\
&=4 - 3c\sqrt[3]{x^2-y}=c^3
\end{align}
From $c^3 + 3c\sqrt[3]{x^2-y} - 2x = 0$ and $c=1$ being a solution we must have $$y= x^2 + \left(\dfrac {2x-1}3\right)^3$$ where $x \equiv 2 \pmod 3$ for $y$ to be an integer.
For positive integer values of $y$ we have $3\sqrt[3]{x^2-y} = 1-2x$. Hence the cubic equation reduces to
$$c^3+(1-2x)c-2x= 0=(c-1)(c^2+c+2x)$$
For $x > \dfrac18$, $c^2+c+2x$ has no solution in the reals. Hence $c=1$, showing that this condition is indeed necessary and sufficient.

One could, of course, obtain the explicit form of $\sqrt[3]{x\pm\sqrt y}$ with our condition, by observing:
$$8 (x \pm \sqrt y) = 1^3 \pm 3\sqrt {\frac {8x-1}3} +3 \sqrt {\frac {8x-1}3}^2 \pm \sqrt {\frac {8x-1}3}^3 = \left(1 \pm \sqrt{\frac {8x-1}3}\right)^3$$
Which gives:
$$\sqrt[3]{x\pm\sqrt y} = \frac{1 \pm \sqrt{\frac {8x-1}3}}2$$
and indeed sums up to $1$, as desired.
A: 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).
A: Note that
$$\sqrt[3]{2-\sqrt{5}}+\sqrt[3]{2+\sqrt{5}}=1$$
is a special case $(m=1)$ of the identity
$$s=\sqrt[3]{\left(3m-1\right) -\sqrt{ (8m-3)m^2}}
 + \sqrt[3]{\left(3m-1\right) +\sqrt{ (8m-3)m^2}} =1
$$
It can be verified by taking cubic power of both sides, i.e.
$$s^3-3(1-2m)s +2-6m=(s-1)(s^2+s+6m-2)=0$$
which leads to $s=1$. Then, write the identity in the form
$$\sqrt[3]{x-\sqrt{y}}+\sqrt[3]{x+\sqrt{y}}=1$$
$$x= 3m-1,\>\>\>\>\>y=(8m-3)m^2$$
$x$ and $y$ are both positive integer for $m=1,2,3...$. and there are infinite possibilities. The first few examples are listed below.
\begin{align}
\sqrt[3]{2-\sqrt{5}} + \sqrt[3]{2+\sqrt{5} } & =1 ,\>\>\>m=1\\
\sqrt[3]{5-\sqrt{52}} + \sqrt[3]{5+\sqrt{52} } & =1 ,\>\>\> m=2\\
\sqrt[3]{8-\sqrt{189}} + \sqrt[3]{8+\sqrt{189} } & = 1 ,\>\>\>m=3\\
 \sqrt[3]{11-\sqrt{464}} + \sqrt[3]{11+\sqrt{464} } & = 1 ,\>\>\>m=4\\
\sqrt[3]{14-\sqrt{925}} + \sqrt[3]{14+\sqrt{925} } & = 1 ,\>\>\>m=5\\
...&= 1,\>\>\>...
\end{align}
