Cho a^3 + b^3 + c^3 = 3abc. Tính B= (1+a/b)(1+b/c)(1+c/a)

* Hướng dẫn giải

Ta có: $a^3+b^3+c=3abc$

$\begin{array}{l}{\left(a+b\right)}^3-3ab\left(a+b\right)+c^3-3abc=0\\ {\left(a+b\right)}^3+c^3-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left[{\left(a+b\right)}^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\\ \left(a+b+c\right)\left[{\left(a+b\right)}^2+\left(a+b\right)c+c^2-3abc\right]=0\\ Do{\left(a+b\right)}^2+\left(a+b\right)c+c^2-3abc>0\\ \Rightarrow a+b+c=0\Rightarrow \left\{\begin{array}{l}a+b=-c\\ a+c=-b\\ b+c=-a\end{array}\right.\end{array}$

Ta có: $B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{a}.\frac{a+c}{b}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1$

Vậy B = -1