Question

[Trần Khuyên]

c/m vs mọi số nguyên x, y, z thì

P=(x+y+z)^3-(y+z-x)^3-(x+z-y)^3-(x+y-z)^3 chia hết cho 24

Answer 1

[Lê Nhật Khôi]

Đặt y+z-x=a

      x+z-y=b

      x+y-z=c

Ta thấy a+b+c=y+z-x+x+z-y+x+y-z=x+y+z

Ta có: \(P=\left(a+b+c\right)^3-a^3-b^3-c^3\)

\(=\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2-a^3-b^3-c^3\)

\(=3a^2b+3ab^2+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

\(=3ab\left(a+b\right)+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)

\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)

\(=3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)

\(=3\cdot2z\cdot2y\cdot2x\)

\(=24xyz⋮24\)

Vậy P chia hết cho 24