This is a problem from Discrete Mathematics and its Applications
Here is my work so far
It's similar to this other question I had Next step to take to reach the contradiction?.
I am assuming that I what I am trying to prove is false, that is there is a positive perfect cube less than 1000 that is the sum of the cubes of two positive integers. From here I need to take logical steps to show that assuming that what I am trying to prove is false leads to a contradiction(false no matter what is passed in).
The first algebraic step I took was saying that if s^3 < 1000, cubing root both sides, s < 10. The next logical step i took was saying that if a^3 + b ^3 = s ^ 3, a^3 + b^3 < 1000 as well. If a ^ 3 + b ^ 3 < 1000, I implied that a ^ 3 < 1000 and b ^ 3 < 1000 because if either component >= 1000, the result will also be >= 1000. Going off of that, repeating(recursion) the step I did for s^3 < 1000. I got that b < 10 and a < 10.
In the end i have three intervals 0 < s < 10, 0 < a < 10, and 0 < b < 10. I don't where to go from here though. I think it be too brute force to test every s from 1 to 9 and showing that there is no a^3 and b^3 that will add to it.
What's the next step I should take to reach the contradiction that would be more efficient/not as exhaustive?