Cho ab+bc+ca=0 và a,b,c khác 0 CM (3a^3-bc)/a^3bc=1/b^3+1/c^3

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1. Giải thích các bước giải:

    Ta có:

   \[\begin{array}{l}
   ab + bc + ca = 0 \Leftrightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \Leftrightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\\
   \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} – \frac{3}{{abc}} = {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} – 3.\frac{1}{a}.\frac{1}{b}.\left( {\frac{1}{a} + \frac{1}{b}} \right) + \frac{1}{{{c^3}}} – \frac{3}{{abc}}\\
    = \left[ {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^3} + \frac{1}{{{c^3}}}} \right] – \frac{3}{{ab}}\left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\\
    = {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} + \frac{1}{{{c^3}}}\\
    = \left( {\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right)\left( {{{\left( {\frac{1}{a} + \frac{1}{b}} \right)}^2} – \left( {\frac{1}{a} + \frac{1}{b}} \right).\frac{1}{c} + \frac{1}{{{c^2}}}} \right)\\
    = 0\\
    \Rightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}} – \frac{1}{{{a^3}}}\\
    \Leftrightarrow \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{{3{a^2} – bc}}{{{a^3}bc}}
   \end{array}\]

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