Question

Cho \(\left(a+b+c\right)^2=a^2+b^2+c^2\)và \(a,b,c\ne0\)cmr \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)

Answer 1

Có: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow ab+bc+ac=0\)

\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)(do a,b,c khác 0)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Suy ra: \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)

\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\)

\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{3}{abc}\)(vì \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\))

Vậy...........