Question
If (x+1/x)2 = 3 Then the value of
x96 + x66 + x42 + x18 + x12 + 1 isSolution
(x+1/x)2 = 3 x + 1/x = √3 On cubing x3 + 1/x3 + 3√3 = 3√3 x3 + 1/x3 = 0 x6 + 1 = 0 x6 = -1 Now; x96 + x66 + x42 + x18 + x12 + 1 (x6 )16 + (x6 )11 + (x6 )7 + (x6 )3 + (x6 )2 + 1 1 + (-1) + (-1) + (-1) + 1 + 1 1 - 1 - 1 - 1 + 1 + 1 = 0
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