bài 2$A=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}=\frac{a^2-ab^2+b^2c-bc^2+c^2a-ca^2}{abc}$
PTĐT thành nhân tử => $A=\frac{(a-b)(a-c)(b-c)}{abc}$
$B=\frac c{a-b}+\frac a{b-c}+\frac c{c-a}=\frac{bc^2-abc-c^3+ac^2+a^2c-a^3-abc+a^2b+ab^2-b^3-abc+b^2}{(a-b)(b-c)(c-a)}=\frac{a^3+b^3+c^3-a^2b-ab^2-b^2c-bc^2-a^2c-ac^2+3abc}{(a-b)(a-c)(b-c)}$
Tử thức $=(a+b)^3+c^3-4a^2b-4ab^2-b^c-bc^2-a^c-ac^2+3abc=(a+b+c)((a+b)^2-(a+b)c+c^2)-4ab(a+b)-bc(b+c)-ac(a+c)+3abc$
Vì $a+b+c=0$=>$\begin{cases}a+b=-c \\a+c=-b\\ b+c=-a \end{cases}$
=>$B=\frac{9abc}{(a-b)(a-c)(b-c)}$
=> $P=A.B=9$