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.
 A: 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 $p$.
A: Without using a specific formula, you can also construct a solution as follows:
$$\begin{align}&\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p},\thinspace x,y\in\mathbb Z^{+}; p\in\mathbb P\\
\implies &x+y+3\sqrt [3]{xy}\left(\sqrt[3]{x}+\sqrt[3]{y}\right)=p\\
\implies &x+y+3\sqrt[3]{pxy}=p\\
\implies &27 pxy=(p-x-y)^3 \\
\implies &p\mid (p-x-y)^3\\
\implies &p\mid p-x-y \\
\implies &p≤p-x-y ,\thinspace p>x+y≥2 \\
\implies &x+y≤0,~\text{A contradiction.}\end{align}$$
A: 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
$\sqrt[3]{xyp}, \sqrt[3]{x^2p}, \sqrt[3]{y^2p}, \sqrt[3]{xp^2}, \sqrt[3]{yp^2}, \sqrt[3]{xy^2}, \sqrt[3]{x^2y} \in \mathbb{Z}$
A: 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 $x_1,y_1.$
