Concept:Let the common value be k, sum the logs to find abc=1, then evaluate the required expression.Explanation:Let b−cloga=c−alogb=a−blogc=k.Then loga=k(b−c), logb=k(c−a), logc=k(a−b).Adding: loga+logb+logc=k[(b−c)+(c−a)+(a−b)]=k⋅0=0.Thus log(abc)=0, so abc=1.Now consider aabbcc. Take its logarithm:log(aabbcc)=aloga+blogb+clogc=a⋅k(b−c)+b⋅k(c−a)+c⋅k(a−b)=k[a(b−c)+b(c−a)+c(a−b)]=k[ab−ac+bc−ab+ac−bc]=k⋅0=0.Therefore log(aabbcc)=0, i.e., aabbcc=1.Answer:aa⋅bbcc=1.