Consider the expression. (√3+i)n+(√3−i)n The term √3+i can be written as, √3+i=−2ωi And, √3−i=2ω2i Here, ω is the complex cube root of unity. Therefore, (√3+i)8+(√3−i)8=(−2ωi)8+(2ω2i)8 =28(ω8+ω16) =28(ω2+ω) =−28 Solve further, (√3+i)8+(√3−i)8=−28 =−256