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Question

x3+x2x+22 if x=512i

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Solution

x=512i
=512i×1+2i1+2i=5+10i1+4=5+10i5=1+2i
x3+x2x+22
(1+2i)3+(1+2i)212i+22
1+(8i)3+3(2i)+3(4i2)+1+4i2+4i2i+21
18i+6i12+14+2i+21
12i11+2i+21
=21+111=2211=11

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