Given that ‌x6=1⇒x6−1=0 . . . (i) ‌⇒(x−1)(x5+x4+x3+x2+x+1)=0 ‌⇒‌‌x5+x4+x3+x2+x+1=0 ‌[∵‌ roots are non-real ‌] ‌ Since α is a root of the equation (i) ∴‌α5+α4+α3+α2+α+1=0 ⇒α5+α3+α+1=−(α4+α2) ⇒α5+α3+α+1=−α2(α2+1) ‌‌