Đáp án:
\(abc \vdots 3\)
Giải thích các bước giải:
\(\begin{array}{l} {a^3} + {b^3} + {c^3} = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ac} \right) + 3abc\\ = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac - 3ac - 3bc - 3ac} \right) + 3abc\\ = \left( {a + b + c} \right)\left[ {{{\left( {a + b + c} \right)}^2} - 3\left( {ab + bc + ac} \right)} \right] + 3abc\\ =(a+b+c)^3-3(a+b+c)(ab+bc+ac)+3abc\\ Do:{a^3} + {b^3} + {c^3} \vdots 9\\ \to {a^3} + {b^3} + {c^3} \vdots 3\\ Có: 3(a+b+c)(ab+bc+ac)\vdots 3, 3abc\vdots 3\\ \Rightarrow (a + b + c)^3 \vdots 3\\ a,b,c\in Z \to a + b + c \vdots 3\\ \Rightarrow (a+b+c)^3 \vdots 3\\ a + b + c \vdots 3 \to 3(a+b+c)(ab+bc+ac)\vdots 9\\ {a^3} + {b^3} + {c^3} \vdots 9\\ \to 3abc \vdots 9\\ \to abc \vdots 3 \end{array}\)
