The sum of the cubes of two different prime numbers is 5256. Is an simple way to solve the problem? 

The sum of the cubes of two different prime numbers is 5256. What are the two primes?

Here is what I did: assume the two numbers are $x$ and $y$, then I have $x^3+y^3=5256$. I can try x and y, but I don't think it is the most effective method. Can anyone help me?

We might as well take $x \ge y$.  Then $18 \gt \sqrt[3]{5256} \gt 17$, so $x$ cannot be greater than $17$.  $x$ has to be at least $\sqrt[3]{5256/2} \gt 13$ so we know $x=17$ as it is the only prime in the range.  Now $y=\sqrt[3]{5256-17^3}=7$ and we are done without checking any cases.
