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  • If a ,b belong to C are distinct root of the Equation x^2-X+1=0 then a^101+b^107=?

If a ,b belong to C are distinct root of the Equation x^2-X+1=0 then a^101+b^107=?

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Solution: 

a, b are the root of \ x^2-x+1=0

\ herefore a=-omega ;and ; b=-omega ^2

Where omega is the non real cube root unity

So,\ a^101+b^107

\ Rightarrow (-omega )^101+(-omega ^2)^107=-[omega ^2+omega ]=-[-1]=1

\ (As ; 1+omega +omega ^2=0 );And ;omega ^3=1

Posted by

Deependra Verma

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