Question

cho \(a^3+b^3+c^3=3abc\)

tính : \(A=\left(\dfrac{a}{b}+1\right)\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)

Answer 1

Từ \(a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a+b+c=0\) hoặc \(a=b;b=c;c=a\) (bn tự chứng minh)

Với \(a+b+c=0\Rightarrow a+b=-c;b+c=-a;a+c=-b\)

Ta có: \(A=\left(\dfrac{a}{b}+1\right).\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)

\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)

Với \(a=b;b=c;c=a\)

\(\Rightarrow A=\dfrac{a+b}{b}.\dfrac{b+c}{c}+\dfrac{c+a}{a}=\dfrac{2b}{b}.\dfrac{2c}{c}.\dfrac{2a}{a}=8\)