+729 Find one ordered pair (x,y) of real numbers such that x + y = 6 and x^3 + y^3 = 162 - (x^2 + y^2).
+129771 x + y = 6 square both sides
x^2 + 2xy + y^2 = 36
x^2 + y^2 = 36 -2xy
x^3 + y^3 = 162 - (x^2 + y^2)
(x + y) (x^2 + y^2 - xy) = 162 - (x^2 + y^2)
6 (x^2 + y^2 - xy) = 162 -(x^2 + y^2)
6 (36 - 2xy -xy) = 162 - (36 -2xy)
216 - 18xy = 126 + 2xy
90 = 20xy
xy = 9/2
y = 9/(2x)
x + y = 6
x + 9/(2x) = 6
2x^2 + 9 = 12x
2x^2 -12x + 9 = 0
x = [12 + sqrt [ 144 - 72] ] / 4 = [ 12 + 6sqrt2] / 4 = 3 + 3/sqrt 2
y = 3 - 3/sqrt 2
