Question

Cho a\(^3\)\(+b^3+c^3=3abc\). Tính giá trị biểu thức:

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

Answer 1

a³ + b³ + c³ = 3abc

<=> a³ + b³ + c³ - 3abc = 0

<=> (a + b + c)(a² + b² + c² - ab - bc - ca) = 0

<=> (a + b + c)[(a - b)² + (b - c)² + (c - a)²] = 0

<=> \(\left[{}\begin{matrix}a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\\\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\Leftrightarrow a=b=c\end{matrix}\right.\)

TH1: a + b + c = 0

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

TH2: a = b = c

A = 2.2.2 = 8