Question

Cho  \(a,b,c\in Z\) để \(\left(a-b\right)\left(b-c\right)\left(c-a\right)=a+b+c\)

CMR: \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3⋮81\)

Answer 1

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

Để tổng trên chia hết cho 81 thì \(\left(a-b\right)\left(b-c\right)\left(c-a\right)⋮27\)

Mà \(a+b+c=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Bài toán trở thành: Cho \(x+y+z=\left(x-y\right)\left(y-z\right)\left(z-x\right)\). CMR: \(x+y+z⋮27\) - Hoc24