This question comes from the Quest for 750 Quant Homework Set. I have a question about the answer explanation and if anyone has any suggestions on how to solve this problem quickly (possibly by plugging in values?). (The answer is C)

If a and b are distinct integers and their product is not equal to zero, is a > b?

(1) (a(^3)b – b(^3)a)/(a(^3)b + b(^3)a – 2a(^2)b(^2)) < 0

(2) b < 0

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(1) Rephrased: (a + b)/(a – b) < 0

The answer explanation then instructs the following computations:
So statement (1) simplifies to (a + b)/(a – b) < 0. If a – b is positive (in other words, if a > b), we can simplify even further:

IF a > b:
a + b < a – b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b < -b
2b < 0
b < 0

This tells us that b < 0 whenever a > b. The converse of this statement is also true. Whenever b < 0, a must be greater than b for the inequality to hold (try plugging sample values into (a + b)/(a – b) < 0 to verify that when b is negative, an a value smaller than b fails to satisfy the inequality).

However, if a – b is negative (in other words, if a < b), we must flip the sign when multiplying through by a – b:
IF a < b:
a + b > a – b (MY MAIN CONCERN: HOW IS IT POSSIBLE TO MULTIPLY (a-b) by 0?)
b > -b
2b > 0
b > 0

This tells us that b > 0 whenever a < b. The converse of this statement is also true. Whenever b > 0, a must be less than b for the inequality to hold (try plugging sample values into (a + b)/(a – b) < 0 to verify that when b is positive, a values greater than b fail to satisfy the inequality).

um yeah, wow, i don't so much follow that one either.

here's how i'd do it. it's essentially the same work, but approximately in reverse of the order presented there.

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statement (1):

(a + b)/(a - b) < 0
therefore, a + b and a - b have opposite signs. we can divide this statement into 2 cases.

CASE 1: a + b is positive and a - b is negative
a - b is negative --> immediately know a < b
also, in this case, a + b > a - b, so therefore b > -b, so therefore b is positive.
that's all we know, though; we know nothing about the sign of a. (note that this case works for (a, b) = (2, 4) but also (-2, 4))

CASE 2: a + b is negative and a - b is positive
a - b is positive --> immediately know a > b
in this case, a + b < a - b, so therefore b < -b, so therefore b is negative.
again, that's all we know. (this case works for (a, b) = (2, -4) but also (-2, -4))

this is insufficient, because there's a case in which a < b and a case in which a > b.

statement 2:
obviously insufficient

together:
this has to be CASE 2 above, so therefore a > b.
sufficient.

hth.