Question

若w=-1/2-根号3/2i,求w^5+w^4+w^3+w^2+w+1=?

Answer 1

由w=-1/2-根号3/2i可知w^2+w+1=0,

因此w^5+w^4+w^3+w^2+w+1=W^3(w^2+w+1)+w^2+w+1=(W^3+1)(w^2+w+1)=0