Given x + y = 3, x ^ 2 + y ^ 2 + 3xy = 4, find the value of x ^ 3Y + XY ^ 3

Given x + y = 3, x ^ 2 + y ^ 2 + 3xy = 4, find the value of x ^ 3Y + XY ^ 3

x^2+y^2+3xy=4 (1)
Because x + y = 3, we get (x + y) ^ 2 = x ^ 2 + y ^ 2 + 2XY = 9
That is x ^ 2 + y ^ 2 + 2XY = 9 (2)
(1) (2) get xy = - 5
From (1), x ^ 2 + y ^ 2 = 4-3xy
So x ^ 3Y + XY ^ 3 = XY (x ^ 2 + y ^ 2)
=xy(4-3xy)
=(-5)[4-3*(-5)]
=95

Given x + y = 3, X2 + y2-3xy = 4. Find the value of the following formula: (1) XY; & nbsp; & nbsp; (2) x3y + XY3

(1) ∵ x + y = 3, ∵ (x + y) 2 = 9, ∵ x2 + Y2 + 2XY = 9, ∵ x2 + y2 = 9-2xy, substituting x2 + y2-3xy = 4, ∵ 9-2xy-3xy = 4, the solution is xy = 1. (2) ∵ x2 + y2-3xy = 4, xy = 1, ∵ x2 + y2 = 7, and ∵ x3y + XY3 = XY (x2 + Y2), ∵ x3y + XY3 = 1 × 7 = 7