Ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\left(-\dfrac{1}{c}\right)^3\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=-\dfrac{1}{c^3}\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+3.\dfrac{1}{ab}.\left(-\dfrac{1}{c}\right)=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\Leftrightarrow abc.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3abc}{abc}\Leftrightarrow\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=3\Leftrightarrow P=3\)Vậy khi \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) thì P=3
