If $a = 29$, $b = 24$, $c = 27$, the value of $a^3 + b^3 + c^3 - 3abc$ is

### Concept & Formula When the values of $a$, $b$, and $c$ are large but relatively close to each other, computing their cubes directly is inefficient. Instead, use the specialized form of the cubic identity involving squares of differences: $$a^3 + b^3 + c^3 - 3abc = \frac{1}{2}(a + b + c)[(a - b)^2 + (b - c)^2 + (c - a)^2]$$ ### Step-by-Step Solution * **Given:** $a = 29$, $b = 24$, $c = 27$. * First, calculate the sum of the variables: $$a + b + c = 29 + 24 + 27 = 80$$ * Next, calculate the differences between each pair of variables: $$a - b = 29 - 24 = 5$$ $$b - c = 24 - 27 = -3$$ $$c - a = 27 - 29 = -2$$ * Square these differences: $$(a - b)^2 = 5^2 = 25$$ $$(b - c)^2 = (-3)^2 = 9$$ $$(c - a)^2 = (-2)^2 = 4$$ * Sum the squares of the differences: $$25 + 9 + 4 = 38$$ * Plug everything into the specialized formula: $$\frac{1}{2} \times (80) \times (38)$$ * Simplify the calculation: $$= 40 \times 38 = 1520$$ ### Exam Strategy & Shortcut Using the difference-of-squares formula $\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2]$ is the ultimate shortcut here. It reduces large multi-digit cube calculations to basic addition and single-digit squaring. ### Common Pitfall The most common trap is attempting brute-force calculation ($29^3 + 24^3 + 27^3 - \dots$). This will consume several minutes and likely result in an arithmetic mistake. ### Final Answer Therefore, the correct answer is **1520**.