If $p, q, r$ are all real numbers then $(p - q)^3 + (q - r)^3 + (r - p)^3$ is equal to

## Concept & Formula This problem relies on a very specific and highly tested algebraic identity involving the sum of three cubes. The conditional algebraic identity states that if $a + b + c = 0$, then: $$a^3 + b^3 + c^3 = 3abc$$ ## Step-by-Step Solution * **Given:** The expression to evaluate is $(p - q)^3 + (q - r)^3 + (r - p)^3$. * **Substitution:** Let's define three new variables to map this to our standard identity: $$a = p - q$$ $$b = q - r$$ $$c = r - p$$ * **Calculation:** Check the sum of these new variables ($a + b + c$). $$a + b + c = (p - q) + (q - r) + (r - p)$$ $$a + b + c = p - q + q - r + r - p = 0$$ * **Deduction:** Since the condition $a + b + c = 0$ is perfectly satisfied, we can directly apply the rule $a^3 + b^3 + c^3 = 3abc$. * Substitute $a, b,$ and $c$ back into the formula: $$(p - q)^3 + (q - r)^3 + (r - p)^3 = 3(p - q)(q - r)(r - p)$$ ## Exam Strategy & Shortcut Whenever you see a cyclic sum of cubes like $(x - y)^3 + (y - z)^3 + (z - x)^3$ in a competitive exam, you should instantly recognize it as a zero-sum condition. You don't need to write out the proof; memorize that the result will always be $3 \times \text{(product of the terms)}$. ## Common Pitfall A frequent mistake is choosing $0$ (Option d) as the answer. Students correctly identify that the inner terms sum to $0$, but mistakenly assume that the sum of their *cubes* will also be $0$. Always remember the $3abc$ outcome. ## Final Answer Therefore, the correct answer is $3(p - q)(q - r)(r - p)$.