Ta có:

a³ + b³ + c³ = 3abc

⇔ (a + b)³ – 3ab(a + b) + c³ – abc = 0

⇔ (a + b + c)[(a + b)² + (a + b)c + c²] – 3ab(a + b + c) = 0

⇔ (a + b + c) [(a + b)² + (a + b)c + c² – 3abc] = 0

Do (a + b)² + (a + b)c + c² – 3abc > 0 nên a + b + c = 0

Suy ra:

$\left\{\begin{array}{l}a+b=-c\\ a+c=-b\\ b+c=-a\end{array}\right.$

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.