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Solution to puzzle 9: Reciprocals and cubes

1/x + 1/y = −1(1)
x3 + y3 = 4(2)

(1) implies x + y = −xy
(2) implies (x + y)3 − 3xy(x + y) = 4
Hence  −(xy)3 + 3(xy)2 − 4 = 0

By inspection, xy = −1 is a solution of this cubic equation.
Factorizing, we have (xy + 1)(xy − 2)2 = 0.
Hence xy = −1, x + y = 1, or xy = 2, x + y = −2. 

If xy = −1 and x + y = 1, then x, y are roots of the quadratic equation u2 − u − 1 = 0.
(Consider the sum and product of the roots of (u − A)(u − B) = u2 − (A + B)u + AB = 0.)
Hence u = (1 ± root 5)/2. 

If xy = 2 and x + y = −2, then x, y are roots of u2 + 2u + 2 = 0.
This has complex roots: u = −1 ± i.

Therefore the real solutions are  x = (1 ± root 5)/2,  y = (1 minus or plus root 5)/2.

Source: Original

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