\(P=2\left(x^3+y^3\right)-3\left(x^2+y^2\right)\)
\(=2.\left(x+y\right)\left(x^2-xy+y^2\right)-3x^2-3y^2\)
\(=2\left(x^2-xy+y^2\right).1-3x^2-3y^2\)
\(=2x^2-2xy+2y^2-3x^2-3y^2=-x^2-2xy-y^2\)
\(=-\left(x^2+2xy+y^2\right)=-\left(x+y\right)^2=-1\)
Vậy P=-1