JEE Mains 25-July-2022 Shift 1 Solved Paper

When, x5=1
then x5−1=0
⇒(x−1)(x4+x3+x2+x+1)=0
Given, x4+x3+x2+x+1=0 has roots α,β,γ and 8 .
∴ Roots of x5−1=0 are 1,α,β,γ and 8
We know, Sum of p‌th ‌ power of n‌th ‌ roots of unity =0. (If p is not multiple of n ) or n (If p is multiple of n )
∴ Here, Sum of p‌th ‌ power of n‌th ‌ roots of unity
Here, p=2021, which is not multiple of 5 .
∴12021+α2021+β2021+γ2021+82021=0
⇒α2021+β2021+γ2021+82021=−1