Đáp án:
\[M = 0\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{a^3} + {b^3} + {c^3}\\
= \left( {{a^3} + 3{a^2}b + 3a{b^2} + {b^3}} \right) + {c^3} - \left( {3{a^2}b + 3a{b^2}} \right)\\
= {\left( {a + b} \right)^3} + {c^3} - 3ab\left( {a + b} \right)\\
= \left[ {\left( {a + b} \right) + c} \right].\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right).c + {c^2}} \right] - 3ab\left( {a + b} \right)\\
= \left( {a + b + c} \right).\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right)c + {c^2}} \right] - 3ab\left( {a + b} \right)\\
a + b + c = 0 \Rightarrow {a^3} + {b^3} + {c^3} = 0.\left[ {{{\left( {a + b} \right)}^2} - \left( {a + b} \right)c + {c^2}} \right] - 3ab.\left( { - c} \right) = 3abc\\
*)\\
x + y + z = 10 \Rightarrow \left( {x + 6} \right) + \left( {y - 7} \right) + \left( {z - 9} \right) = \left( {x + y + z} \right) - 10 = 0\\
\Rightarrow {\left( {x + 6} \right)^3} + {\left( {y - 7} \right)^3} + {\left( {z - 9} \right)^3} = 3.\left( {x + 6} \right).\left( {y - 7} \right).\left( {z - 9} \right)\\
\Rightarrow 3.\left( {x + 6} \right).\left( {y - 7} \right).\left( {z - 9} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 6\\
y = 7\\
z = 9
\end{array} \right.\\
TH1:\,\,\,x = - 6 \Rightarrow y + z = 16 \Leftrightarrow y - 7 = - \left( {z - 9} \right)\\
M = {\left( {x + 6} \right)^{2019}} + {\left( {y - 7} \right)^{2019}} + {\left( {z - 9} \right)^{2019}}\\
= {\left( { - 6 + 6} \right)^{2019}} + {\left[ { - \left( {z - 9} \right)} \right]^{2019}} + {\left( {z - 9} \right)^{2019}}\\
= 0\\
TH2:\,\,\,y = 7 \Rightarrow x + z = 3 \Leftrightarrow x + 6 = - \left( {z - 9} \right)\\
M = {\left( {x + 6} \right)^{2019}} + {\left( {y - 7} \right)^{2019}} + {\left( {z - 9} \right)^{2019}}\\
= {\left[ { - \left( {z - 9} \right)} \right]^{2019}} + {\left( {7 - 7} \right)^{2019}} + {\left( {z - 9} \right)^{2019}}\\
= 0\\
TH3:\,\,\,z = 9 \Rightarrow x + y = 1 \Leftrightarrow x + 6 = - \left( {y - 7} \right)\\
M = {\left( {x + 6} \right)^{2019}} + {\left( {y - 7} \right)^{2019}} + {\left( {z - 9} \right)^{2019}}\\
= {\left[ { - \left( {y - 7} \right)} \right]^{2019}} + {\left( {y - 7} \right)^{2019}} + {\left( {9 - 9} \right)^{2019}}\\
= 0
\end{array}\)
Vậy \(M = 0\)
