Question

Cho a,b,c\(\ne\)0 và \(a^3+b^3+c^3\)=3abc

Tính giá trị của biểu thức \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)

Answer 1

Thay a³+b³=(a+b)³-3ab(a+b) vào giả thiết ta có:

(a+b)³-3ab(a+b)+c³-3abc=0

<=> [(a+b)+c].\(\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)-3ab(a+b+c)=0

<=> (a+b+c) (a²+b²+c²-ab-bc+c²-3ab)=0

<=> (a+b+c)(a²+b²+c²-ab-bc-ca)=0

\(\Leftrightarrow\orbr{\begin{cases}a+b+c=0\\a^2+b^2+c^2-ab-bc-ca=0\end{cases}}\)

Nếu a+b+c=0

\(\Rightarrow A=\frac{b+a}{b}\cdot\frac{c+b}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}\Rightarrow A=-1\)

Nếu \(a^2+b^2+c^2-ab-bc-ca=0\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=> a=b=c

Khi đó \(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)