Ta có:
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\) \(=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2-\left(a-c\right)^2\right]\)
\(=\left(a-c\right)\left[\left(a-b\right)\left(a-2b+c\right)+\left(b-a\right)\left(a+b-2c\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left[\left(a-2b+c\right)-\left(a+b-2c\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(3c-3b\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)=3k\)
Vì \(3k⋮3k\) nên \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3⋮3k\)
