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