Theo giả thiết, ta có:
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow\) \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)
\(\Rightarrow\) \(2\left(ab+bc+ac\right)=0\)
\(\Rightarrow\) \(ab+bc+ac=0\)
Vì \(a,b,c\ne0\) nên \(\frac{ab+bc+ac}{abc}=0\), tức là \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) \(\left(1\right)\)
Từ \(\left(1\right)\) \(\Rightarrow\) \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\) \(\left(2\right)\)
\(\Leftrightarrow\) \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+3.\frac{1}{a}.\frac{1}{b}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)
\(\Leftrightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\Leftrightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\) (do \(\left(2\right)\) )
