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Solve the system x + y = 2 and x³ + y³ = 56 for real x and y.

1. x = 4, y = -2 or x = -2, y = 4
2. x = 3, y = -1 or x = -1, y = 3
3. x = 1, y = 1
4. x = 5, y = -3 or x = -3, y = 5

Correct answer: x = 4, y = -2 or x = -2, y = 4

Solution

Let s = x + y = 2 and p = xy. Then x³ + y³ = s³ - 3ps = 8 - 6p = 56, giving p = -8. The numbers x, y are roots of t² - 2t - 8 = 0, i.e. t = 4 or t = -2.

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