BTGmoderatorDC wrote: ↑Sun Jul 01, 2018 12:24 am
If xy ≠0, is x^3 + y^3 > 0 ?
(1) x + y > 0
(2) xy > 0
Solution:
Question Stem Analysis:
We need to determine whether x^3 + y^3 > 0, given that neither x nor y is 0.
Statement One Only:
Since x + y > 0, either both x and y are positive or one of them is positive and the other is negative. If both x and y are positive, then obviously x^3 + y^3 > 0. If one of them is positive and the other is negative, without loss of generality, we can let x be positive and y be negative. In order for x + y > 0, x must be “more” positive than y is negative (in other words, |x| > |y|) . In that case, x^3 will also be “more” positive than y^3 (in other words, |x^3| > |y^3|). Therefore, x^3 + y^3 > 0. Statement one alone is sufficient.
(Alternatively, recall that x^3 + y^3 = (x + y)(x^2 - xy + y^2). Again, if both x and y are positive, then obviously x^3 + y^3 > 0. Now, if one of the values is positive and the other is negative, then xy is negative and hence -xy is positive. Since x^2 + y^2 is positive, x^2 - xy + y^2 will be positive. Therefore, given the fact that x + y is positive, we see that (x + y)(x^2 - xy + y^2) will be positive, i.e., x^3 + y^3 is positive.)
Statement Two Only:
Since xy > 0, either both x and y are positive or both are negative. If both x and y are positive, then obviously x^3 + y^3 > 0. However, if both are negative, then x^3 + y^3 < 0. Statement two is not sufficient.
Answer: A
