Ta có :\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3\cdot\frac{1}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=3\cdot\frac{1}{abc}\)
( Do \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) )
Khi đó : \(P=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)