Bài 7 : Cho $a,b,c\ne 0$ và ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$. Chứng minh rằng $\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=\frac{3}{abc}$

Giải

Ta có : ${{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

$\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\,(ab+bc+ca)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$

$\Leftrightarrow ab+bc+ca=0$

$\Leftrightarrow \frac{ab+bc+ca}{abc}=0$

$\Leftrightarrow \frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=0$

$\Leftrightarrow \frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0$ hay $\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$

Đặt $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$

$\Rightarrow x+y+z=0$ $\Leftrightarrow x+y=-z$$\Leftrightarrow -{{(x+y)}^{3}}={{z}^{3}}$

Lại có : ${{x}^{3}}+{{y}^{3}}+{{z}^{3}}$

$={{x}^{3}}+{{y}^{3}}-{{(x+y)}^{3}}$

$={{x}^{3}}+{{y}^{3}}-\left( {{x}^{3}}+3xy(x+y)+{{y}^{3}} \right)$

$={{x}^{3}}+{{y}^{3}}-{{x}^{3}}-{{y}^{3}}-3xy(x+y)$

$=-3xy(x+y)$

$=-3xy.(-z)$

$=3xyz$

Thay $\frac{1}{a}=x$; $\frac{1}{b}=y$; $\frac{1}{c}=z$ ta được

${{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz$

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

$\Leftrightarrow \frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=\frac{3}{abc}$

Vậy $\frac{1}{{{a}^{3}}}+\frac{1}{{{b}^{3}}}+\frac{1}{{{c}^{3}}}=\frac{3}{abc}$