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