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Question

The sum of 162th power of the roots of the equation x32x2+2x1=0 is

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Solution

Let roots of x32x2+2x1=0 be α,β,γ.
(x31)(2x22x)=0
(x1)(x2x+1)=0
x=1,ω,ω2

Now, α162+β162+γ162
=1+(ω)162+(ω2)162
=1+(ω3)54+(ω3)108=3

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