Ta có:
$\left\{\begin{array}{l}x^3+y^3+3xyz=z^3\\(2x+2y)^2=z^3\end{array}\right.$
$\Leftrightarrow \left\{\begin{array}{l}(x+y-z)[\dfrac{1}{2}(x-y)^2+\dfrac{1}{2}(x+z)^2+\dfrac{1}{2}(y+z)^2]=0\\4(x+y)^2=z^3\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x+y=z\\4(x+y)^2=z^3\end{array}\right.\\\left\{\begin{array}{l}x=y=-z\\4(x+y)^2=z^3\end{array}\right.\end{array}\right.$
$\Leftrightarrow \left[\begin{array}{l}\left\{\begin{array}{l}x+y=0\\z=0\end{array}\right.\\\left\{\begin{array}{l}x+y=4\\z=4\end{array}\right.\\x=y=z=0\\\left\{\begin{array}{l}x=y=-16\\z=16\end{array}\right.\\\end{array}\right.$