Quantity I: $$ \frac{(2a^3 + 2b^3)(a-b)}{(8a^2 - 9ab + 2b^2 + 7ab - 6a^2)(a^2 - b^2)} $$ Quantity II: Value of P – 2Q if Q < (–1) < P, Where P and Q are integers.

### Concept & Algebraic Identity Evaluation To compare Quantity I and Quantity II, we must independently evaluate or find the range for each. For Quantity I, we apply standard algebraic factorizations. For Quantity II, we use inequality logic for integers. Formulas used: $$ a^3 + b^3 = (a+b)(a^2 - ab + b^2) $$ $$ a^2 - b^2 = (a-b)(a+b) $$ ### Step-by-Step Solution 1. **Evaluate Quantity I:** Numerator: $(2a^3 + 2b^3)(a-b)$ Factor out the 2: $2(a^3 + b^3)(a-b)$ Expand sum of cubes: $2(a+b)(a^2 - ab + b^2)(a-b)$ Denominator: $(8a^2 - 9ab + 2b^2 + 7ab - 6a^2)(a^2 - b^2)$ Simplify the first polynomial: $8a^2 - 6a^2 - 9ab + 7ab + 2b^2 = 2a^2 - 2ab + 2b^2 = 2(a^2 - ab + b^2)$ Expand the second term: $(a^2 - b^2) = (a-b)(a+b)$ Combine denominator: $2(a^2 - ab + b^2)(a-b)(a+b)$ Divide Numerator by Denominator: $$ \frac{2(a+b)(a^2 - ab + b^2)(a-b)}{2(a^2 - ab + b^2)(a-b)(a+b)} = 1 $$ So, **Quantity I = 1**. 2. **Evaluate Quantity II:** We are given $Q < -1 < P$, and both are integers. The maximum possible value for integer $Q$ is $-2$. The minimum possible value for integer $P$ is $0$. We want the minimum value of $P - 2Q$: Substitute the minimum value of $P$ and maximum value of $Q$ (since it's being subtracted): Minimum $P - 2Q = 0 - 2(-2) = 0 + 4 = 4$. Thus, **Quantity II $\geq 4$**. 3. **Compare Quantities:** Quantity I ($1$) is less than Quantity II (which is at least $4$). ### Exam Strategy & Shortcut For Quantity I, recognizing that $2a^2 - 2ab + 2b^2$ is exactly the missing piece of the $a^3+b^3$ expansion allows for immediate cancellation without writing out all the factors. For Quantity II, immediately testing boundary integer values ($P=0$, $Q=-2$) proves Quantity II is positive and $\geq 4$, immediately solving the comparison. ### Common Pitfall A common mistake is failing to properly combine the like terms in the denominator of Quantity I, making it look unfactorable, or misinterpreting the integer constraints in Quantity II (e.g., using fractional values like $-1.5$). ### Final Answer Therefore, the correct answer is **Quantity I < Quantity II**.