Need help with a number theory proof

Is there a way to show that if there exists two natural numbers a and b coprime to each other, and a third prime p coprime to both a and b, that a^3 + b^3 cannot equal to some integer power of p with the exception of 1^3 + 1^3 = 2 and 1^3 + 2^3 = 3^2.

$\displaystyle \begin{align*} \gcd(a,b) &= 1 \\ \gcd(a,p) &= 1 \\ \gcd(b,p) &= 1 \\ a^3 + b^3 &\ne p^k \hspace{4mm} \mbox{for some } a,b,p,k \in \mathbb{N} \text{ and p is prime}\end{align*}$

Re: Need help with a number theory proof

Alright, I got it

$\displaystyle \begin{align*}a^3 + b^3 &= (a + b)(a^2 - ab + b^2) \\ a^2 - ab + b^2 &= (a + b)^2 - 3ab \\ a^3 + b^3 &= (a+b)((a+b)^2 - 3ab)\\&= p^k\end{align*}$

Assuming p,a,b are all relatively prime, and p is prime, then, it must be the case that an expression of the form $\displaystyle (a+b)*x = p^n \iff p|(a+b) \wedge p|x$.

If $\displaystyle p|(a+b)$, then obviously $\displaystyle p|(a+b)^2$, hence $\displaystyle p|((a+b)^2-3ab) \iff p|3ab$. This means that either $\displaystyle p|a$ or $\displaystyle p|b$, but this contradicts the assumption that a,b,p are all mutually coprime. Hence, there cannot be any solutions to my equation under these constraints. $\displaystyle \square$