\(\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left[\left(a+b\right)+c\right]^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3-a^3-b^3-c^3\)

\(=a^3+3a^2b+3ab^2+b^3+3c\left(a^2+2ab+b^2\right)+3ac^2+3bc^2-a^3-b^3\)

\(=3a^2b+3ab^2+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)

\(=3\left(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2+2abc\right)\)

\(=3\left[\left(a^2b+ab^2\right)+\left(a^2c+abc\right)+\left(ac^2+bc^2\right)+\left(b^2c+abc\right)\right]\)

\(=3\left[ab\left(a+b\right)+ac\left(a+b\right)+c^2\left(a+b\right)+bc\left(a+b\right)\right]\)

\(=3\left(a+b\right)\left(ab+ac+c^2+bc\right)\)

\(=3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=3\left(a+b\right)\left(b+c\right)\left(c+b\right)\)

\(P=\frac{x-y}{x+y}\)\(\Rightarrow P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}\)

Ta có: \(2x^2+2y^2=5xy\left(1\right)\)

\(\Leftrightarrow2x^2-4xy+2y^2=xy\)

\(\Leftrightarrow2\left(x-y\right)^2=xy\)

\(\Leftrightarrow\left(x-y\right)^2=\frac{xy}{2}\)(2)

Từ \(\left(1\right)\Rightarrow2x^2+4xy+2y^2=9xy\)

\(\Leftrightarrow2\left(x+y\right)^2=9xy\)

\(\Leftrightarrow\left(x+y\right)^2=\frac{9xy}{2}\)(3)

Thay (2)và (3) vào \(P^2\)ta được:

\(P^2=\frac{\left(x-y\right)^2}{\left(x+y\right)^2}=\frac{xy}{2}.\frac{2}{9xy}=\frac{1}{9}\)

Nhận thấy \(y>x>0\Rightarrow x-y< 0\Rightarrow\frac{x-y}{x+y}< 0\Rightarrow P< 0\)

\(\Rightarrow P=\frac{-1}{3}\)