Ta có: \(P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}\)
=> \(\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}=abc\left(\dfrac{1+1+1}{a^3+b^3+c^3}\right)\)
Mà ta lại có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)
\(\dfrac{\Rightarrow1}{a^3}+\dfrac{1}{b^3}=\left(\dfrac{1+1}{a+b}\right)^3-3\dfrac{1}{a}.\dfrac{1}{b}\left(\dfrac{1+1}{a+b}\right)\)
\(\Rightarrow-\dfrac{1^3}{c}+\dfrac{3}{abc}\)
\(\Rightarrow\dfrac{1+1+1}{a^3+b^3+c^3}=\dfrac{3}{abc}\)
\(\dfrac{\Rightarrow bc}{c^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}=3\)
Vậy : \(P=\dfrac{ab}{c^2}+\dfrac{bc}{a^2}+\dfrac{ca}{b^2}\) = 3
