Có $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0$
$→\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}$
$→(\dfrac{1}{a}+\dfrac{1}{b})^3=(-\dfrac{1}{c})^3$
$→\dfrac{1}{a^3}+\dfrac{1}{b^3}+3.\dfrac{1}{a}.\dfrac{1}{b}.(\dfrac{1}{a}+\dfrac{1}{b})=-\dfrac{1}{c^3}$
$→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{a}.\dfrac{1}{b}.(\dfrac{1}{a}+\dfrac{1}{b})$
Mà $\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}$
$→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=-3.\dfrac{1}{a}.\dfrac{1}{b}.-\dfrac{1}{c}$
$→\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=3.\dfrac{1}{a}.\dfrac{1}{b}.\dfrac{1}{c}=3.\dfrac{1}{abc}$
Có: $P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ac}{b^2}$
$=\dfrac{abc}{c^3}+\dfrac{abc}{a^3}+\dfrac{abc}{b^3}$
$=abc.(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3})$
Mà $\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=3.\dfrac{1}{abc}$
$=abc.3.\dfrac{1}{abc}$
$=3$
