Ta có: \(x^3+y^3+\frac{1}{3^3}-3xy.\frac{1}{3}=0\)
<=> \(\left(x+y+\frac{1}{3}\right)\left(x^2+y^2+\frac{1}{9}-xy-\frac{1}{3}x-\frac{1}{3}y\right)=0\)
<=> \(\orbr{\begin{cases}x+y+\frac{1}{3}=0\left(1\right)\\x^2+y^2+\frac{1}{9}-xy-\frac{1}{3}x-\frac{1}{3}y=0\left(2\right)\end{cases}}\)
(1) <=> \(x+y=-\frac{1}{3}\)loại vì x > 0 ; y >0
( 2) <=> \(\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(x-y\right)^2=0\)
vì \(\left(x-\frac{1}{3}\right)^2\ge0;\left(y-\frac{1}{3}\right)^2\ge0;\left(x-y\right)^2\ge0\)với mọi x, y
nên \(\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(x-y\right)^2\ge0\)với mọi x, y
Do đó: \(\left(x-\frac{1}{3}\right)^2+\left(y-\frac{1}{3}\right)^2+\left(x-y\right)^2=0\)
<=> \(x=y=\frac{1}{3}\)
Làm tiếp:
Với \(x=y=\frac{1}{3}\)=> \(x+y=\frac{2}{3}\) thế vào P
ta có: \(P=\left(\frac{2}{3}+\frac{1}{3}\right)^3-\frac{3}{2}.\frac{2}{3}+2016=2016\)