can you explain why from q^3<p^3 follows q<p ?

thanks for sharing this general rule. you saved me a lot of time i would have wasted on testing numbers.

An algebraic way of explaining this, for those curious:

p² > q² = |p|² > |q|², so

p² > q² implies |p| > |q|, but NOT p > q

(For instance, if p = -3 and q = 2, p² > q² and |p| > |q|, but p is not > q.)

For p³ > q³, it's a little more complicated. We have:

p³ > q³, or
p³ - q³ > 0, or
(p - q) * (p² + pq + q²) > 0

(that's how you factor the difference of cubes ... don't worry about it on the GMAT, I haven't seen it tested there yet)

This tells us that either (p - q) and (p² + pq + q²) are both positive OR (p - q) and (p² + pq + q²) are both negative.

Let's assume that it's the second case. If (p - q) and (p² + pq + q²) are both negative, then (p² + pq + q²) < 0. But if (p² + pq + q²) < 0, then (p² + q²) < -pq. Since (p² + q²) CAN'T be negative, -pq must be positive, which means that exactly one of p and q is negative.

Since we're assuming (p - q) < 0, we have p < q, so p is the negative one and q the positive. But if q is positive and p is negative, we can't have p³ > q³ (how can a positive be greater than a negative!?) This contradicts our original statement (that p³ > q³), making it impossible.

Hence (p - q) and (p² + pq + q²) are both positive. Since (p - q) is positive, p > q.

Woof: too much work! This is why you want to try numbers on test day

what if p = 1 and q = 2? 1 cubed is LESS than 2 cubed and in this case p < q and not p>q?