Đáp án:
$M = 0$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
{a^3} + {b^3} + {c^3} = 3abc\\
\Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = 0\\
\Leftrightarrow {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc = 0\\
\Leftrightarrow {\left( {a + b} \right)^3} + {c^3} - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{{\left( {a + b} \right)}^2} - \left( {a + b} \right)c + {c^2}} \right) - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab - ac - bc} \right) - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2ac - 2bc} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{{\left( {a - b} \right)}^2} + {{\left( {a - c} \right)}^2} + {{\left( {b - c} \right)}^2}} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
a + b + c = 0\\
{\left( {a - b} \right)^2} + {\left( {a - c} \right)^2} + {\left( {b - c} \right)^2} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a + b + c = 0\\
{\left( {a - b} \right)^2} = {\left( {a - c} \right)^2} = {\left( {b - c} \right)^2} = 0\left( {Do:{{\left( {a - b} \right)}^2} + {{\left( {a - c} \right)}^2} + {{\left( {b - c} \right)}^2} \ge 0,\forall a,b,c} \right)
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
a + b + c = 0\\
a = b = c\left( l, do:a\ne b\ne c \right)
\end{array} \right.\\
\Leftrightarrow a + b + c = 0\\
\Rightarrow \left\{ \begin{array}{l}
a + b = - c\\
a + c = - b\\
b + c = - a
\end{array} \right.\\
\Rightarrow \left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right) = \left( { - a} \right)\left( { - b} \right)\left( { - c} \right) = - abc\\
\Rightarrow \left( {a + b} \right)\left( {a + c} \right)\left( {b + c} \right) + abc = 0
\end{array}$
$ \Rightarrow M = 0$
Vậy $M = 0$
