$\begin{align}
& {{x}^{2}}\left( x+3 \right)+{{y}^{2}}\left( y+5 \right)-\left( x+y \right)\left( {{x}^{2}}-xy+{{y}^{2}} \right)=0 \\
& \Rightarrow {{x}^{3}}+3{{x}^{2}}+{{y}^{3}}+5{{y}^{2}}-\left( {{x}^{3}}+{{y}^{3}} \right)=0 \\
& \Rightarrow 3{{x}^{2}}+5{{y}^{2}}=0 \\
\end{align}$
${{x}^{2}}\ge 0\,\,\,\forall x\in \mathbb{R},\,\,{{y}^{2}}\ge 0\,\,\,\forall y\in \mathbb{R}\Rightarrow 3{{x}^{2}}+5{{y}^{2}}\ge 0\,\,\,\,\,\,\forall x,\,\,y\in \mathbb{R}$
Dấu “=” xảy ra khi và chỉ khi $\left\{ \begin{align}
& {{x}^{2}}=0 \\
& {{y}^{2}}=0 \\
\end{align} \right.\Rightarrow x=y=0$
Vậy bộ $\left( x,\,\,y \right)$ thỏa mãn là $\left( 0,\,\,0 \right)$.