P=\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
=\(\left(\dfrac{a+b}{b}\right)\left(\dfrac{b+c}{c}\right)\left(\dfrac{a+c}{a}\right)\)
Vi a+b+c=0 nen a+b=-c , b+c=-a , a+c=-b nen :
P=\(\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
= -1