+1348 Find one ordered pair (x,y) of real numbers such that x + y = 6 and x^3 + y^3 = 162 - (x^2 + y^2).
+129895 x^3 + y^3 = 162 - (x^2 + y^2)
(x + y) (x^2 - xy + y^2) = 162 - (x^2 + y^2
6( x^2 + y^2) - 6xy = 162 - (x^2 + y^2)
7 (x^2 + y^2 + 2xy) -20xy = 162
7 ( x + y)^2 - 20xy = 162
7 (6)^2 - 20xy =162
90 = 20xy
9/2 = xy
(9/2)/x = y
x + (9/2)/x = 6
x^2 + (9/2) = 6x
x^2 - 6x = -9/2
x^2 - 6x + 9 = -9/2 + 9
(x - 3)^2 = 9/2
x -3 = sqrt(9/2) or x - 3 = -sqrt (9/2)
x = 3 + 3/sqrt (2) x = 3 - 3/sqrt (2)
(x, y) = ( 3 + 3/sqrt2 , 3- 3/sqrt 2) and ( 3 - 3/sqrt2 , 3 + 3/sqrt 2)
