Đặt \(a+b-c=x;b+c-a=y;c+a-b=z\)
\(\Rightarrow x+y+z=a+b-c+b+c-a+c+a-b\)
\(=a+b+c\)
Thay \(x;y;z;x+y+z\) vào M, ta được:
\(M=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+z^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-x^3-y^3-z^3\)
\(=x^3+y^3+z^3-x^3-y^3-z^3+3xy\left(x+y\right)+3z\left(x+y\right)\left(x+y+z\right)\)\(=3\left(x+y\right)\left[xy+z\left(x+y+z\right)\right]\)
\(=3\left(x+y\right)\left(xy+xz+zy+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(a+b-c+c+a-b\right)\)
\(=3.2b.2c.2a=24abc\)
Vì \(24abc⋮24\forall a,b,c\) nên \(M⋮24\)
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