Решите систему уравнений

$\left \{ {{9(x^4+y^4)=17(x+y)^2} \atop {3xy=-2(x+y)}} \right. \\$
(x+y)=a xy=b

(x+y)² = x²+y² + 2xy; a² = x²+y² + 2b; a² - 2b = x²+y²

(x²+y²)² = x⁴+y⁴ + 2x²y²; (a² - 2b)² = x⁴+y⁴ + 2b²; a⁴ -4a²b + 4b² = x⁴+y⁴ + 2b²;
= x⁴+y⁴

$\left \{ {{9(a^4 - 4aвb + 2bв)=17aв} \atop {3b=-2a}} \right. \\ \left \{ {{9a^4 - 36baв - 17aв + 18bв= 0 } \atop {3b=-2a}} \right. \\ \left \{ {{9a^4 - (36b + 17)aв + 18bв= 0 } \atop {3b=-2a}} \right. \\ \left \{ {{9a^4 - (-24a + 17)aв + 8aв= 0 } \atop {3b=-2a}} \right. \\ \left \{ {{9a^4 + 24a^3 - 9aв = 0 } \atop {3b=-2a}} \right. \\ \left \{ {{aв(9aв + 24a - 9) = 0 } \atop {3b=-2a}} \right. \\ \left \{ {{aв(3aв + 8a - 3) = 0 } \atop {3b=-2a}} \right.$

$\left \{ {{(3aв + 8a - 3) = 0 } \atop {3b=-2a}} \right. \left \{ {{a=0} \atop {b=0}} \right. \\ \left \{ {{ \left \ [ {{a=-3} \atop {x= \frac13 }} \right. } \atop {3b=-2a}} \right.РРРРРРР \left \{ {{x=0} \atop {y=0}} \right. \\ \left \{ {{ a=-3 } \atop {b=2}} \left \{ {{a= \frac13} \atop {b= -\frac29}} \right. \right.\left \{ {{x=0} \atop {y=0}} \right. \\ \left \{ {{ x=-2 } \atop {y=-1}} \left \{ {{x,y= \frac{1б\sqrt5}{3} } }} \right. \right.\left \{ {{x=0} \atop {y=0}} \right.$