Chứng minh rằng: `a^3+b^3+c^3` chia hết cho 6 khi và chỉ khi a+b+c chia hết cho 6

$\begin{array}{l}\quad a^3 + b^3 + c^3\\ = a^3 - a + b^3 - b + c^3 - c + a +b + c\\ = a(a^2 - 1) + b(b^2 - 1) + c(c^2 - 1) + a +b + c\\ = (a+1).a.(a-1) + (b+1).b.(b-1) + (c+1).c.(c-1) + a+ b+ c\\ Ta\,\,có:\\ \quad \begin{cases}(a+1).a.(a-1)\quad \vdots \quad 6\\(b+1).b.(b-1)\quad \vdots \quad 6\\ (c+1).c.(c-1) \quad \vdots \quad 6\end{cases}\\ \Rightarrow (a+1).a.(a-1) + (b+1).b.(b-1) + (c+1).c.(c-1) \quad \vdots \quad 6\\Do\,\,đó:\\ (a+1)a(a-1) + (b+1)b(b-1) + (c+1)c(c-1) + a+ b+ c \quad \vdots \quad 6\\ \Leftrightarrow a + b+ c \quad \vdots \quad 6\\ Hay \,\,a^3 +b^3 + c^3 \quad \vdots \quad 6\Leftrightarrow a +b + c \quad \vdots \quad 6\end{array}$