Positive numbers a and b, with a > b, satisfy a² + b² = 6ab. What is the value of (a³ + b³)/(a³ − b³)?
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Answer & Explanation
Correct answer
B
Do not solve for a and b individually. The relation a² + b² = 6ab gives a/b = 3 + 2√2, an ugly irrational that makes cubing intractable. Instead, use a² + b² = 6ab to build the symmetric pairs (a + b)² and (a − b)².
Add 2ab to both sides of a² + b² = 6ab: (a + b)² = a² + 2ab + b² = 6ab + 2ab = 8ab. Subtract 2ab instead: (a − b)² = a² − 2ab + b² = 6ab − 2ab = 4ab. Dividing these, (a + b)²/(a − b)² = 8ab/4ab = 2. Since a > b and both are positive, a + b and a − b are both positive, so (a + b)/(a − b) = √2.
Now factor the cubes using the same relation. The sum of cubes is (a + b)(a² − ab + b²). Since a² + b² = 6ab, the second factor becomes 6ab − ab = 5ab. The difference of cubes is (a − b)(a² + ab + b²) = (a − b)(6ab + ab) = (a − b)(7ab). So the ratio (a³ + b³)/(a³ − b³) equals [(a + b)(5ab)] / [(a − b)(7ab)]. The ab cancels, leaving (5/7) × (a + b)/(a − b) = (5/7) × √2, which is 5√2/7.
As a numerical check, take b = 1 and a ≈ 5.828 (a solution of a² + 1 = 6a). The ratio (a³ + b³)/(a³ − b³) comes out to about 1.010, which matches 5√2/7 ≈ 1.010. So the answer is B.