Quantity I. ((2a³ + 2b³)(a − b)) / ((6a² − 7ab + 2b² + 5ab − 4a²)(a² − b²))…

Concept

A two-quantity comparison is decided by reducing each side to its actual value or value-range, then comparing. Two identities drive Quantity I: the sum of cubes a³ + b³ = (a + b)(a² − ab + b²) and the difference of squares a² − b² = (a + b)(a − b). For Quantity II, a linear expression in integers is bounded by pushing each variable to its extreme allowed value.

Application — Quantity I

1. Numerator: 2a³ + 2b³ = 2(a³ + b³) = 2(a + b)(a² − ab + b²); multiplied by (a − b) gives 2(a + b)(a − b)(a² − ab + b²).

2. Denominator bracket: 6a² − 7ab + 2b² + 5ab − 4a² = (6a² − 4a²) + (−7ab + 5ab) + 2b² = 2a² − 2ab + 2b² = 2(a² − ab + b²).

3. Other denominator factor: a² − b² = (a + b)(a − b). So the full denominator is 2(a² − ab + b²)(a + b)(a − b).

4. Cancel the common factors 2, (a + b), (a − b) and (a² − ab + b²) from numerator and denominator. Quantity I = 1.

Application — Quantity II

1. Q is an integer with Q < −1, so the largest Q can be is −2.

2. P is an integer with P > −1, so the smallest P can be is 0.

3. P − 2Q is smallest when P is smallest and −2Q is smallest. −2Q is smallest when Q is largest (Q = −2), giving −2(−2) = 4.

4. Minimum of Quantity II = 0 − 2(−2) = 0 + 4 = 4. As P can grow and Q can fall further, Quantity II ranges over 4, 5, 6, … and upward, i.e. Quantity II ≥ 4.

Cross-check & result

Quantity I is the fixed value 1, while Quantity II is at least 4 for every admissible pair of integers. Hence 1 < Quantity II always, so Quantity I < Quantity II.