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Solve the system of equations: x³ + y³ = 7 and xy(x + y) = -2.

1. (x, y) = (-1, 2) or (2, -1)
2. (x, y) = (1, 2) or (2, 1)
3. (x, y) = (-1, -2) or (-2, -1)
4. (x, y) = (1, -2) or (-2, 1)

Correct answer: (x, y) = (-1, 2) or (2, -1)

Solution

Let s = x + y, p = xy. Then x³ + y³ = s³ - 3ps = 7 and ps = -2. Substitute ps = -2: s³ - 3(-2) = 7 -> s³ + 6 = 7 -> s³ = 1 -> s = 1. Then p = -2/s = -2. So x + y = 1, xy = -2; x, y are roots of u² - u - 2 = 0 = (u - 2)(u + 1), giving u = 2 or u = -1. Thus (x, y) = (2, -1) or (-1, 2).

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