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Question

If t2+t+1=0, then the value of the expression (t+1t)+(t2+1t2)++(t27+1t27) equals

A
0
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B
1
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C
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D
none
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Solution

The correct option is A 0
Let S=(t+1t)+(t2+1t2)++(t27+1t27)
=(t+t2+............+t27)+(1t+1t2+..........+1t27)
=(t(1t27)1t)+(1t(1(1t)27)11t)
=(t+1t27)(1t27)1t
Now since t2+t+1=0t=ω,ω2
and we know ω3n=1ω27=1
Hence S=0

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