Đáp án:
\[N = \frac{1}{3}\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
{x^3} + {y^3} = {\left( {x + y} \right)^3} - 3xy\left( {x + y} \right)\\
{a^3} + {b^3} + {c^3} = 3abc\\
\Leftrightarrow {a^3} + {b^3} + {c^3} - 3abc = 0\\
\Leftrightarrow {\left( {a + b} \right)^3} - 3ab\left( {a + b} \right) + {c^3} - 3abc = 0\\
\Leftrightarrow \left[ {{{\left( {a + b} \right)}^3} + {c^3}} \right] - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow {\left( {a + b + c} \right)^3} - 3\left( {a + b} \right)c.\left( {a + b + c} \right) - 3ab\left( {a + b + c} \right) = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left[ {{{\left( {a + b + c} \right)}^2} - 3\left( {a + b} \right)c - 3ab} \right] = 0\\
\Leftrightarrow \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} - ab - bc - ca = 0\\
\Leftrightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca\\
N = \frac{{{a^2} + {b^2} + {c^2}}}{{{{\left( {a + b + c} \right)}^2}}} = \frac{{{a^2} + {b^2} + {c^2}}}{{{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right)}} = \frac{{{a^2} + {b^2} + {c^2}}}{{3\left( {{a^2} + {b^2} + {c^2}} \right)}} = \frac{1}{3}
\end{array}\)
