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