Решить систему уравнений : х³+3х²у=20, 3ху²+у³=7

Сложим два уравнения:

x³ + 3x²y + 3xy² + y³ = 27

(x+y)³ = 3³

x + y = 3

x = 3 - y

3(3-y)y² + y³ = 7
9y² - 3y³ + y³ = 7
2y³ - 9y² + 7 = 0
2y²(y-1) - 7y(y-1) - 7(y - 1) = 0
(y-1)(2y² - 7y - 7) = 0
y1 = 1
2y² - 7y - 7 = 0
D = 49 + 56 = 105
y2 = (7 - √105)/4
y3 = (7 + √105)/4

x1 = 2
x2 = 3 - (7 - √105)/4 = (5 + √105)/4
x3 = 3 - (7 + √105)/4 = (5 - √105)/4

ответ: ((5 - √105)/4; (7 + √105)/4); ((5 + √105)/4; (7 - √105)/4); (2; 1)