• Factorize $a^3+b^3$.
• $p$ can not be the multiplication of two integers greater than $1$. So, $a^2-ab+b^2$ is bounded.
• Use the bound to get the value of $(a-b)^2$
•  Get the value of $p$

Assume that the prime is $p$.

$p^2 = a^3+b^3 = (a+b) (a^2-ab+b^2)$

Case 1:

$a+b=1$ and $a^2-ab+b^2 = p^2$

Given that $a,b\geq1$, $a+b\geq2$ which contradicts the case.

Case 2:

$a+b=p^2$ and $a^2-ab+b^2 = 1\implies (a-b)^2 = 1-ab$

But $(a-b)^2\geq 0$ and $a-b = 1-ab<1$

As $(a-b)$ is an integer, only possible value of $(a-b)^2$ is 0 which implies $a=b$

So, $2a^2-a^2 = 1 \implies a =b= 1 \implies p^2 = 2$ which also doesn't have any integer solution.

Case 3:

$a+b=p\implies a=p-b$ 

And $a^2-ab+b^2=p\implies (p-b)^2-(p-b)b+b^2=p\implies  p^2-3pb+3b^2=p$

$\implies p(3b+1-p) = 3b^2 \implies p\mid 3b^2$

Either, $p\mid 3 \implies p=3$ which is $p^2=9$.

9 can be expressed as $a^3+b^3$ as $1^3+2^3$ and $2^3+1^3$ and these are the only ways to do that.

So, $(a,b,p) = (1,2,3)$ and $(2,1,3)$ work.

Or, $p\mid b$. Similarly, $p\mid a$.

Assume that $a = pA$ and $b = pB$

So, $p^2 = a^3+b^3 = p^3(A^3+B^3) \implies p(A^3+B^3) = 1$

But, that is not possible for any value of $p$.

So, the only solutions are 

$(a,b,p) = (1,2,3)$ and $(2,1,3)$

Key points:

• Factorize $a^3+b^3$.
• $p$ can not be the multiplication of two integers greater than 1. So, there are 3 cases.
• The cases are $a+b=1,~a+b=p^2$ and $a+b=p$
• $a+b\geq 2$ eliminates the first case. For the second case, $a^2-ab+b^2$ can be bounded.
• Solve the third case using $a=p-b$ and expanding $a$ in the other equation.