What is the absolute difference between the cubes of two

What is the absolute difference between the cubes of two different non-negative integers?

Given: \(x\) and \(y\) are different non-negative integers. Question: \(|x^3-y^3|=?\)

(1) One of the integers is 2 greater than the other integer --> \(|x-y|=2\). There exist infinitely many such \(x\) and \(y\), thus we'll have infinitely many different values of \(|x^3-y^3|\). For example consider \(x=2\), \(y=0\) and \(x=3\), \(y=1\). Not sufficient.

(2) The square of the sum of the integers is 49 greater than the product of the integers --> \((x+y)^2=xy+49\) --> \(x^2+xy+y^2=49\). Now, we can not simply say that this statement is not sufficient (as done in the above post), because we have some constraints on \(x\) and \(y\), namely that they are non-negative integers. Taking this into account, from \(x^2+xy+y^2=49\) we can say that both \(x\) and \(y\) are integers from 0 to 7, inclusive. By trial and error we can easily find that only two pairs satisfy this equation: (7,0) and (5,3), in any order. Not sufficient.

Notice that if we were told that \(x\) and \(y\) are positive integers (instead of non-negative integers) then this statement would be sufficient, since only one pair would remain: (5,3).

(1)+(2) Since from (1) \(|x-y|=2\) then from (2) only one pair is valid: (5,3). Sufficient.

Answer: C.

Hope it's clear.