ta có $x+y+x=0 \Rightarrow x^{3}+y^{3}+z^{3}=3xyz$$ab+bc=ca=0 \Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$ $\Rightarrow \frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}=\frac{3}{abc}$P=$\frac{abc}{a^{3}}+\frac{abc}{b^{3}}+\frac{abc}{c^{2}}=abc.\frac{3}{abc} =3$
ta có $x+y+
z=0 \Rightarrow x^{3}+y^{3}+z^{3}=3xyz$$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=$\frac{abc}{a^{3}}+\frac{abc}{b^{3}}+\frac{abc}{c^{
3}}=abc.\frac{3}{abc} =3$