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展開して整理しています。
(1+2i)³=1+3×(-4)+3×(2i)+(-8i)=-11-2iより
(1+2i)³+a(1+2i)+b=0
⇔-11-2i+a+2ai+b=0
⇔(a+b-11)+(2a-2)i=0

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