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Question

If x=ωω22, then the value of x4+3x3+2x211x6 is (where ω is cube root of unity)

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Solution

Given that x=ωω22
x+2=ωω2
On squaring & rearranging, we get x2+4x+7=0 ....(1)
dividing x4+3x3+2x211x6 with x2+4x+7, we get x4+3x3+2x211x6=(x2+4x+7)(x2x1)+1
Therefore using (1), we get x4+3x3+2x211x6=1

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