Given that $p 1 q 0$, compare the following two quantities:

**Quantity I:** Value of $\frac{p^3 - q^3}{p - q} - 3pq$

**Quantity II:** Value of $\left(1 - \frac{1}{q}\right)$

Determine the correct relationship between Quantity I and Quantity II.

Question

Given that $p > 1 > q > 0$, compare the following two quantities:

**Quantity I:** Value of $\frac{p^3 - q^3}{p - q} - 3pq$

**Quantity II:** Value of $\left(1 - \frac{1}{q}\right)$

Determine the correct relationship between Quantity I and Quantity II.

Accepted Answer

**Solution:** 1. Start by simplifying **Quantity I**: $$\frac{p^3 - q^3}{p - q} - 3pq$$ 2. Recall the factorization formula for difference of cubes: $$p^3 - q^3 = (p - q)(p^2 + pq + q^2)$$ 3. Substitute this into Quantity I: $$\frac{(p - q)(p^2 + pq + q^2)}{p - q} - 3pq = p^2 + pq + q^2 - 3pq$$ 4. Simplify the expression: $$p^2 + pq + q^2 - 3pq = p^2 - 2pq + q^2 = (p - q)^2$$ 5. So, **Quantity I** simplifies to: $$(p - q)^2$$ 6. Now, analyze **Quantity II**: $$1 - \frac{1}{q} = \frac{q - 1}{q}$$ 7. Given the conditions $p > 1 > q > 0$, note that: - Since $q < 1$, $q - 1 < 0$, so Quantity II is negative. - Quantity I is a square of a real number, so $(p - q)^2 \geq 0$. 8. Therefore, Quantity I is always non-negative, and Quantity II is negative. 9. Hence, **Quantity I \geq Quantity II** always holds true. **Answer:** Quantity I \geq Quantity II

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