Ta có:
\(a^3+b^3+c^3\\ =\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2+2ab+2bc+2ca-3ab-3bc-3ca\right)+3abc\)\(=\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]+3abc\)
Vì \(a+b+c⋮3\Rightarrow\)\(\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]⋮3\) (1)
Mà \(3abc⋮3\) (2)
Từ (1) và (2) \(\Rightarrow\text{}\text{}\)\(\left(a+b+c\right)\left[\left(a+b+c\right)^2-3\left(ab+bc+ca\right)\right]+3abc⋮3\)
Hay \(a^3+b^3+c^3⋮3\) (ĐPCM)
