<s>[INST] Prove $p^n\ne a^3+b^3$ where $p\in\mathbb{P}\ge5$, $n,a,b\in\mathbb{N}$ 
Given prime $p\geq5$, prove that $p^n$ can't be represented as sum of two positive cubes for any $n\in\mathbb{N}$. What about $p=2$ and $p=3$?

 [/INST] Let $p\ge 5$, and suppose that there is an $n\ge 1$ such that $p^n$ can be represented as $a^3+b^3$, where $a$ and $b$ are positive.  Then there is a smallest such $n$.  
From the factorization $a^3+b^3=(a+b)(a^2-ab+b^2)$, we conclude that $a+b$ and $a^2-ab+b^2$ are each powers of $p$. Neither can be the $0$-th power, since for positive $a$ and $b$, we have $a^2-ab+b^2=1$ only if $a=b=1$. 
But $3ab=(a+b)^2-(a^2-ab+b^2)$. So $p$ divides $3ab$, and since $p\gt 3$, we conclude that $p$ divides one of $a$ or $b$. But since $p$ divides $a+b$, it follows that $p$ divides both $a$ and $b$. Let $a=pc$, $b=pd$. Then $c^3+d^3=p^{n-3}$. 
This contradicts the minimality of $n$ unless $n-3=0$. But $n-3=0$ is impossible, since $c$ and $d$ are positive. 
The cases $p=2$ and $p=3$ are different, since $1^3+1^3=2$ and $1^3+2^3=3^2$. 
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