Solution to puzzle 9: Reciprocals and cubes

1/x + 1/y = −1	(1)
x3 + y3 = 4	(2)

(1) x + y = −xy
(2) (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 ± )/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 ± )/2,  y = (1 )/2.

Source: Original

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