<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] The fundamental problem is on your “Suppose that…” line. You need to prove that, and it can be a bit of work. The rest of the proof is fine.
A similar more direct argument just cubes both sides to get:
$$x+3\sqrt[3]{x^2y}+ 3\sqrt[3]{xy^2}+y=p$$
Letting $u= 3\sqrt[3]{x^2y},v= 3\sqrt[3]{xy^2}$ then $u+v=p-x-y$ and $uv=9xy.$
This means $u$ is a root of the two integer polynomials:
$$f(w)=w^3-27x^2y\\g(w)=w^2-(p-x-y)w+9xy$$
If $u$ is not an integer, then $g$ must be the minimal polynomial for $u,$ so $g$ must divide $f.$ The only possible factorization is:
$$f(w)=(w-3x)g(w).$$
Then $$0=f(3x)=27x^3-27x^2y=27x^2(x-y),$$ and we get $x=0$ or $x=y.$ But $x>0$ and, if $x=y,$ we’d have $$ \sqrt[3]x+\sqrt[3]y=\sqrt[3]{8x}=\sqrt[3]p$$ or $p=8x,$ contradicting that $p$ is prime.
So $u$ must be an integer, and the same for $v.$
So $x^2y$ and $xy^2$ must be perfect cubes. Then $$\frac xy=\frac{x^2y}{xy^2}=\left(\frac ab\right)^3$$ for positive integers $a,b.$ This means $a^3x=b^3y.$
Then $$\begin{align} a\sqrt[3]p&=a(\sqrt[3]x+\sqrt[3]{y})\\
&=(a+b)\sqrt[3]y\end{align}$$
Cubing both sides:
$$a^3p=(a+b)^3y.\tag1$$
So $p$ must divide $y.$ But then $y\geq p.$
Similarly $x\geq p.$
But then $$\sqrt[3]x+\sqrt[3]y>\sqrt[3]p,$$ a contradiction.

The general question of integer roots to:
$$\sqrt[3]x+\sqrt[3]y=\sqrt[3]{z}$$ works the same, but you have to allow $x=0,y=z$ and $x=y,$ and $z=8x.$
When you get to (1), you have to write $z=z_0z_1^3,$ where $z_0$ has non nontrivial cube divisors.  Then for (1) you get:
$$(az_1)^3z_0=(a+b)^3y.$$ This means $y=z_0y_1^3$ and then, similarly, $x=z_0x_1^3,$ and you get $x_1+y_1=z_1,$ so all solutions can be written:
$$(x,y,z)=\left(z_0x_1^3,z_0y_1^3,z_0(x_1+y_1)^3\right)$$
For all positive cube-free $z_0$ and integers $x_1,y_1.$
</s>