Ta có:
\(x+y+z=x^3+y^3+z^3=1\)
\(\Rightarrow\left(x+y+z\right)^3=\left(x+y\right)^3+y^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+3x^2y+3xy^2+z^3+3\left(x+y\right)z\left(x+y+z\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)\left(x+z\right)=0\)
=> x = -y hoặc y = -z hoặc z = -x
Với x = -y => x + y +z = 1 => z = 1
==" tính M = 1 ghê
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