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Two solid cubes have positive edge lengths. The edge of the larger cube is 3 longer than the edge of the smaller cube, and the volume of the larger cube is 126 greater than the volume of the smaller cube. What is the sum of the area of one face of the larger cube and the area of one face of the smaller cube?
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Correct answer
A
Let the edges be m (larger) and n (smaller), so m − n = 3 and the volume difference m³ − n³ = 126. The face areas are m² and n², so the question asks for m² + n². Do not solve for the edges directly: the roots are irrational, so extracting m and n is a dead end. Instead use two identities, with the product mn as the bridge.
Factor the cube difference: m³ − n³ = (m − n)(m² + mn + n²), so with m − n = 3 we get m² + mn + n² = 126 ÷ 3 = 42. Square the edge difference: (m − n)² = m² − 2mn + n² = 9. Subtracting, the two expressions differ only in the product term, so 42 − 9 = 3mn, giving 3mn = 33 and mn = 11. Then m² + n² = (m² + mn + n²) − mn = 42 − 11 = 31. So the answer is A.
Among the traps: 42 is the intermediate value m² + mn + n², reported if you forget to subtract the product; 33 is the subtraction step 42 − 9 (which equals 3mn), reported before dividing by 3; 75 comes from dropping the (m − n) factor on the cross term in the cube expansion, inflating mn to 33; 53 is (m + n)², the sum of squares plus the cross term, reported without removing that cross term.
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