Đáp án:
$A = 0$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
xy\left( {x + y} \right) + yz\left( {y + z} \right) + zx\left( {z + x} \right) + 2xyz = 0\\
\Leftrightarrow xy\left( {x + y} \right) + \left( {{y^2}z + y{z^2} + {z^2}x + z{x^2} + 2xyz} \right) = 0\\
\Leftrightarrow xy\left( {x + y} \right) + \left( {{z^2}\left( {x + y} \right) + z\left( {{x^2} + 2xy + {y^2}} \right)} \right) = 0\\
\Leftrightarrow xy\left( {x + y} \right) + \left( {{z^2}\left( {x + y} \right) + z{{\left( {x + y} \right)}^2}} \right) = 0\\
\Leftrightarrow xy\left( {x + y} \right) + \left( {x + y} \right)\left( {{z^2} + z\left( {x + y} \right)} \right) = 0\\
\Leftrightarrow \left( {x + y} \right)\left( {xy + {z^2} + zx + zy} \right) = 0\\
\Leftrightarrow \left( {x + y} \right)\left( {x + z} \right)\left( {y + z} \right) = 0
\end{array}$
Mà
$\begin{array}{l}
A = ({x^3} + {y^3})({y^3} + {z^3})({z^3} + {x^3})\\
= \left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)\left( {y + z} \right)\left( {{y^2} - yz + {z^2}} \right)\left( {z + x} \right)\left( {{z^2} - zx + {x^2}} \right)\\
= \left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)\left( {{x^2} - xy + {y^2}} \right)\left( {{y^2} - yz + {z^2}} \right)\left( {{z^2} - zx + {x^2}} \right)\\
= 0
\end{array}$
Vậy $A = 0$
