+1095 Find one ordered pair (x,y) of real numbers such that x + y = 8 and x^3 + y^3 = 200 + x^2 + y^2.
+128732 x^3 + y^3 =
(x + y) ( x^2 + y^2 - xy) = (8) (x^2 + y^2 + 200)
This implies that xy = -200 → y = -200/x
So
x + y = 8
x - 200/x - 8 = 0
x^2 - 8x - 200 = 0
x^2 -8x = 200
x^2 - 8x + 16 = 200 + 16
(x - 4)^2 = sqrt 216 take both roots
x - 4 = sqrt (216)
x = sqrt (216) + 4 or x = -sqrt 216 + 4
x = 4 + 6sqrt(6) or x = 4 -6sqrt (6)
y = 4 - 6sqrt (6) y = 4 + 6sqrt (6)
One solution is (x,y) = ( 4 + 6sqrt(6) , 4 -6sqrt (6) )
