Cho ba số thực a, b, c khác 0 thỏa mãn  $a^3+b^3+c^3=3abc.$ Tính giá trị của biểu thức  $A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right).$

Ta có: $a^3+b^3+c^3-3abc$

$={\left(a+b\right)}^3-3ab\left(a+b\right)+c^3-3abc$

$={\left(a+b\right)}^3+c^3-3ab\left(a+b\right)-3abc$

$={\left(a+b\right)}^3+c^3-3ab\left(a+b+c\right)$

$=\left(a+b+c\right)\left[{\left(a+b\right)}^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)$

$=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)$

Suy ra $a^3+b^3+c^3=3abc$  hay $a^3+b^3+c^3-3abc=0$

Nên $a+b+c=0$  hoặc $a^2+b^2+c^2-ab-bc-ca=0\left(*\right)$

Mặt khác $2\left(a^2+b^2+c^2-ab-bc-ca\right)=2a^2+2b^2+2c^2-2ab-2bc-2ca$

$={\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2$

Do đó $a^2+b^2+c^2-ab-bc-ca=\frac{1}{2}\left[{\left(a-b\right)}^2+{\left(b-c\right)}^2+{\left(c-a\right)}^2\right]\ge 0$  với mọi a, b, c

Nên để (*) xảy ra thì $\left\{\begin{array}{l}{\left(a-b\right)}^2=0\\ {\left(b-c\right)}^2=0\\ {\left(c-a\right)}^2=0\end{array}\right.$ , hay $\left\{\begin{array}{l}a-b=0\\ b-c=0\\ c-a=0\end{array}\right.$  tức $a=b=c$ .

⦁ Trường hợp 1: $a+b+c=0$

Suy ra $a+b=-c;b+c=-a;c+a=-b$

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

⦁ Trường hợp 2: $a=b=c$ thì ta được $A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\mathrm{2.2.2}=8$ .