Theo bài ra ta có
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\\ \Rightarrow \frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\\ \Rightarrow \left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\\ \Rightarrow \frac{1}{a^3}+3.\frac{1}{a^2}.\frac{1}{b}+3.\frac{1}{a}.\frac{1}{b^2}+\frac{1}{b^3}=-\frac{1}{c^3}$
$\Rightarrow \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3.\frac{1}{a}.\frac{1}{b}.\left(\frac{1}{a}+\frac{1}{b}\right)$
Do$\Rightarrow \frac{1}{a}+\frac{1}{b}=-\frac{1}{c}$
$\Rightarrow \frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-3.\frac{1}{a}.\frac{1}{b}.\left(-\frac{1}{c}\right)=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}$
$\Rightarrow \left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)abc=3.\frac{1}{abc}.abc$
$\Leftrightarrow \frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=3$(đpcm)
