Lời giải:
\(A=x^3+y^3+z^3-x-y-z\)
\(A=\left(x^3-x\right)+\left(y^3-y\right)+\left(z^3-z\right)\)
\(A=x\left(x^2-1\right)+y\left(y^2-1\right)+z\left(z^2-1\right)\)
\(A=x\left(x-1\right)\left(x+1\right)+y\left(y-1\right)\left(y+1\right)+z\left(z-1\right)\left(z+1\right)\)
\(A=\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)\)
Ta có:\(\left\{{}\begin{matrix}x-1;x;x+1\\y-1;y;y+1\\z-1;z;z+1\end{matrix}\right.\) là 3 số tự nhiên liên tiếp
Suy ra: \(\left\{{}\begin{matrix}\left(x-1\right)x\left(x+1\right)\\\left(y-1\right)y\left(y+1\right)\\\left(z-1\right)z\left(z+1\right)\end{matrix}\right.\) chia hết cho \(6\)
Hay \(A⋮6\left(đpcm\right)\)
