ta có $x+y+z=0 \Rightarrow x^{3}+y^{3}+z^{3}=3xyz$$ab+bc+ca=0 \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$
$\Rightarrow \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}=\frac{3}{abc}$
P=$\frac{abc}{a^{3}}+\frac{abc}{b^{3}}+\frac{abc}{c^{3}}=abc.\frac{3}{abc} =3$