Cho (a^3) + (b^3) + (c^3) = 3abc và a + b + c khác 0. Tính giá trị biểu thức N = ((a^2) + (b^2) + (c^2))/(a + b + c)^2)

\({a^3} + {b^3} + {c^3} = 3abc\)

\({a^3} + {b^3} + {c^3} - 3abc = 0\)

\({\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc = 0\)

\(\begin{array}{l}\left( {a + b + c} \right)\left[ {{{\left( {a + b} \right)}^2} - c\left( {a + b} \right) + {c^2}} \right]\\ - 3ab\left( {a + b + c} \right) = 0\end{array}\)

\(\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - ac - bc} \right) = 0\)

\({a^2} + {b^2} + {c^2} - ab - ac - bc = 0{\rm{ }}\left( {a + b + c \ne 0} \right)\)

\({a^2} + {b^2} + {c^2} = ab + ac + bc\)

\(2\left( {{a^2} + {b^2} + {c^2}} \right) = 2\left( {ab + ac + bc} \right)\)

\(3\left( {{a^2} + {b^2} + {c^2}} \right) = {\left( {a + b + c} \right)^2}\)

\(\frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {a + b + c} \right)}^2}}} = \frac{1}{3}{\rm{ }}\left( {a + b + c \ne 0} \right)\)

Vậy \(N = \frac{1}{3}\).