\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)
\(\Rightarrow\frac{1}{a+b+c}=\frac{bc+ca+ab}{abc}\)
\(\Rightarrow\left(a+b+c\right)\left(bc+ca+ab\right)=abc\)
\(\Rightarrow abc+a^2c+a^2b+b^2c+abc+ab^2+bc^2+ac^2+abc=abc\)
\(\Rightarrow2abc+a^2c+a^2b+b^2c+ab^2+bc^2+ac^2=0\)
\(\Rightarrow\left(abc+a^2b\right)+\left(ac^2+a^2c\right)+\left(b^2c+b^2a\right)+\left(bc^2+abc\right)=0\)
\(\Rightarrow ab\left(a+c\right)+ac\left(a+c\right)+b^2\left(a+c\right)+bc\left(a+c\right)=0\)
\(\Rightarrow\left(ab+ac+b^2+bc\right)\left(a+c\right)=0\)
\(\Rightarrow\left[\left(ab+ac\right)+\left(b^2+bc\right)\right]\left(a+c\right)=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Do đó trong a , b , c luôn có 2 số đối nhau.
Phần 2 : Do vai trò a , b , c như nhau nên coi \(a=-b\)( Do có 2 số đối nhau)
\(\Rightarrow a^n=-b^n\)(Vì n lẻ )
\(\Rightarrow\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{a^n+b^n}{a^n.b^n}+\frac{1}{c^n}=0+\frac{1}{c^n}=\frac{1}{c^n}\)
\(\frac{1}{a^n+b^n+c^n}=\frac{1}{\left(a^n+b^n\right)+c^n}=\frac{1}{0+c^n}=\frac{1}{c^n}\)
\(\Rightarrow\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}\)
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