The sum of the reciprocals of two real numbers is -1, and the sum of their cubes is 4. What are the numbers?

Level: Senior

Let x and y be the two real numbers. Then
(1) 1/x+1/y = -1;
(2) x^3+y^3 = 4.
From (1),
(3) x+y = -xy.
Using (2) and (3) in the identity (x+y)^3 = x^3+y^3+3xy(x+y),
(x+y)^3 = 4-3(x+y)^2,
(x+y)^3+3(x+y)^2-4 = 0.
It is easy to check that x+y = 1 is a solution. By synthetic division,
(x+y-1)[(x+y)^2+4(x+y)+4] = 0,
(x+y-1)(x+y+2)^2 = 0,
x+y = 1 or -2.
From (3), xy = -1 or 2.
Therefore x and y satisfies the quadratic equation
u^2-u-1 = 0 or u^2+2u+2 = 0,
The roots in the first equation are [1+/-sqrt(5)]/2 and there is no real roots in the second. Hence {x, y} = {[1-sqrt(5)]/2, [1+sqrt(5)]/2}.