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Question : 8 of 160
Marks:
+1,
-0
Solution:
(A) ω1010+ω2020=ω336×3+2+ω673×3+1
‌‌=(ω3)336⋅ω2+(ω3)673⋅ω
‌‌=ω2+ω=−1‌‌[∵ω3=1]
=(ω3)336⋅ω2+(ω3)673⋅ω‌
=ω2+ω=−1‌[∵ω3=1]
‌ (B) ‌(1+ω−ω2)(1−ω+ω2)‌[∵1+ω+ω2=0]
‌‌=(−ω2−ω2)(−ω−ω)
‌‌=(−2ω2)(−2ω)
‌‌=4ω3=4
(C) (2+ω2+ω4)5=(2+ω2+ω)5
=(1+1+ω+ω2)5=(1+0)5=1
(D) (3+5ω+3ω2)3=(3+2ω+3(ω+ω2))3
=(3+2ω−3)3=(2ω)3=8
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