\(x^2+y^2=20\)
\(=>-\left(x^2-y^2\right)=20\)
\(=>x^2-y^2=-20\)
\(=>\left(x+y\right)\left(x-y\right)=-20\)
Mà x+y = 2 ( gt )
=> x-y = -10
*Từ
x-y = -10
x+y = 2
=> x-y+x+y = -10+ 2
=> 2x = -8
=> x = \(-4\)
Mà x+y = 2 => y = 6
Thay x= -4 , y =6 vào \(x^3+y^3\)
Có:
\(x^3+y^3=\left(-4\right)^3+6^3=\left(6-4\right)\left(\left(-4\right)^2-24+6^2\right)\))
\(=2.28=56\)
Vậy x\(^3\)+y\(^3\)=56
