Given, x3−2x2+2x−1=0 i.e. (x3−1)−(2x2−2x)=0 ⇒‌(x−1)(x2+x+1)−2x(x−1)=0 ⇒‌(x−1)(x2+x+1−2x)=0 ⇒‌(x−1)(x2−x+1)=0 ∴‌‌x=1‌ and ‌x=‌
−(−1)±√1−4
2
=‌
1±√3i
2
Then, sum of 162‌th ‌ power of the roots ‌‌=(1)162+(−ω)162+(−ω2)162 ‌‌=1+ω162+ω324 ‌‌=1+(ω3)54+(ω3)108 ‌‌=1+(1)54+(1)108[∵ω3=1] ‌‌=1+1+1=3