Giải thích các bước giải:
$\begin{array}{l} {a^3} + {b^3} + {c^3} = 3abc\\ \Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = 0\\ \Leftrightarrow {(a + b)^3} - 3ab(a + b) + {c^3} - 3abc = 0\\ \Leftrightarrow {(a + b)^3} - 3ab(a + b + c) + {c^3} = 0\\ \Leftrightarrow \left[ {{{(a + b)}^3} + {c^3}} \right] - 3ab(a + b + c) = 0\\ \Leftrightarrow (a + b + c)\left[ {{{(a + b)}^2} + {c^2} - c(a + b)} \right] - 3ab(a + b + c) = 0\\ \Leftrightarrow (a + b + c)\left[ {{{(a + b)}^2} + {c^2} - c(a + b) - 3ab} \right] = 0\\ \Leftrightarrow a + b + c = 0 \end{array}$
$\begin{array}{l} \Rightarrow \left\{ \begin{array}{l} a + b = - c\\ b + c = - a\\ c + a = - b \end{array} \right.\\ \Rightarrow \frac{{a + b}}{c} = \frac{{b + c}}{a} = \frac{{a + c}}{b} = - 1\\ \Rightarrow A = \frac{{a + b}}{c}.\frac{{b + c}}{a}.\frac{{a + c}}{b} = - 1\\ = \frac{{a + b}}{b}.\frac{{b + c}}{c}.\frac{{a + c}}{a} = (1 + \frac{a}{b})(1 + \frac{b}{c})(a + \frac{c}{a}) \end{array}$
