\(x^3+y^3+z^3\)
\(=\left(x+y+z\right).\left(x+y+z\right).\left(x+y+z\right)\)
Mà x + y + z chia hết cho 6
\(\Rightarrow x^3+y^3+z^3⋮6\)
k mik nha!
Xét hiệu :
\(\left(x^3+y^3+z^3\right)-\left(x+y+z\right)\)
\(=\left(x^3-x\right)+\left(y^3-y\right)+\left(z^3-z\right)\)
\(=x\left(x^2-1\right)+y\left(y^2-1\right)+z\left(z^2-1\right)\)
\(=\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)\)
Vì các tích \(\left(x-1\right)x\left(x+1\right);\left(y-1\right)y\left(y+1\right);\left(z-1\right)z\left(z+1\right)\) là tích của 3 số TN liên tiếp
Nên \(\hept{\begin{cases}\left(x-1\right)x\left(x+1\right)⋮6\\\left(y-1\right)y\left(y+1\right)⋮6\\\left(z-1\right)z\left(z+1\right)⋮6\end{cases}}\)\(\Rightarrow\left(x-1\right)x\left(x+1\right)+\left(y-1\right)y\left(y+1\right)+\left(z-1\right)z\left(z+1\right)⋮6\)
Hay \(\left(x^3+y^3+z^3\right)-\left(x+y+z\right)⋮6\)
Mà \(\left(x+y+z\right)⋮6\)(gt) \(\Rightarrow x^3+y^3+z^3⋮6\)(đpcm)