<s>[INST] Are there positive integers $x$, $y$ and prime numbers $p$ so that $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$ I have a solution for this, but I'm not really sure about that:

We have: $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$, multiplying both sides by $\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}$ we get $x+y=\sqrt[3]{p}(\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2})$ so $\sqrt[3]{px^2}-\sqrt[3]{pxy}+\sqrt[3]{py^2}$ is an integer.
Suppose $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are all integers, we can easily see that $p|x$ and $p|y$.
Let $x^2=p^2a^3$, $y^2=p^2b^3$  where $a$ and $b$ are positive integers. Then we can see that $a$ and $b$ also have to be perfect squares since $x=p\sqrt{a^3}$ and $y=p\sqrt{b^3}$ are integers. Since $\gcd(2,3)=1$, we can let $x^2=p^2a'^6$, $y^2=p^2b'^6$, or $x=pa'^3$, $y=pb'^3$ where $a'$ and $b'$ are integers. Subbing that to the original equation we get $\sqrt[3]{pa'^3}+\sqrt[3]{pb'^3}=\sqrt[3]{p}$, so $\sqrt[3]{p}(a'+b')=\sqrt[3]{p}$. But $a'+b'>1$, so $\sqrt[3]{p}(a'+b')>\sqrt[3]{p}$, hence a contradiction.

I can see I made a lot of assumptions here, like "Let $x^2=p^2a^3$, $y^2=p^2b^3$", or assuming all $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are integers. Are there any problems with my work or is it good to go? And moreover, do you have a better solution than this? I appreciate your time and effort for this and thanks a lot in advance.
 [/INST] Suppose that $\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{p}$. Then
\begin{gather*}
(\sqrt[3]{x} + \sqrt[3]{y})^3 = x+y+3\sqrt[3]{x}\sqrt[3]{y}(\sqrt[3]{x} + \sqrt[3]{y}) = p \\
\sqrt[3]{x}\sqrt[3]{y}\sqrt[3]{p} =\sqrt[3]{xyp} = \frac{p-x-y}{3} \in \mathbb{Q}
\end{gather*}
It can be shown that this implies that $\sqrt[3]{xyp} \in \mathbb{Z}$
The, multiplying the original equation by $\sqrt[3]{xp}$ and squaring we have
\begin{gather*}
\sqrt[3]{xp^2} = \sqrt[3]{xyp} +\sqrt[3]{x^2p} = z +\sqrt[3]{x^2p} \\
(\sqrt[3]{xp^2})^2 = p\sqrt[3]{x^2p} =z(\sqrt[3]{xp^2}) +xp
\end{gather*}
This is equivalent to a system of 2 equations. Let $a=\sqrt[3]{xp^2}, b=\sqrt[3]{x^2p}$
\begin{align*}
a - b &= z \\
za - pb &= -xp 
\end{align*}
There are 2 cases. If the system is indeterminate, then $z=p=-x$, contradiction. If not, then $a, b \in \mathbb{Q}$, so $a, b \in \mathbb{Z}$. Then also $\sqrt[3]{y^2p} = \frac{(\sqrt[3]{xyp})^2}{\sqrt[3]{x^2p}} \in \mathbb{Q}$ and $\sqrt[3]{yp^2} = \frac{(\sqrt[3]{xyp})(\sqrt[3]{xp^2})}{\sqrt[3]{x^2p}} \in \mathbb{Q}$. A similiar reasoning also leads to $\sqrt[3]{xy^2}, \sqrt[3]{x^2y} \in \mathbb{Q}$
That means that
$\sqrt[3]{xyp}, \sqrt[3]{x^2p}, \sqrt[3]{y^2p}, \sqrt[3]{xp^2}, \sqrt[3]{yp^2}, \sqrt[3]{xy^2}, \sqrt[3]{x^2y} \in \mathbb{Z}$
</s>