<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] Recall the algebraic identity:
$$a^3+b^3+c^3 - 3abc = \frac12(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
Whenever $a+b+c = 0$, we have $a^3+b^3+c^3 = 3abc$.
Substitute $a,b,c$ by $\sqrt[3]{p}, -\sqrt[3]{x}, -\sqrt[3]{y}$ and let $K = p - x - y$, we find
$$K = p - x - y = 3\sqrt[3]{pxy} > 0$$
Since $x,y > 0$, $K$ is a positive integer smaller than $p$.
Taking cube on the equality on the left and using the condition $p$ is a prime, we have
$$K^3 = 27pxy \implies p | K^3 \implies p | K$$
This contradicts with above fact that $K$ is a positive integer smaller than $p$.
As a result, equation $\sqrt[3]{x} + \sqrt[3]{y} = \sqrt[3]{p}$ has not positive integer solutions for any prime $p$.
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