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December 14th, 2013, 09:51 PM  #1 
Newbie Joined: Dec 2013 Posts: 3 Thanks: 0  Cube of a polynomial
Tomorrow is my in house final defence and today i have to add some extra things plz help me. a(x)= (x^96+X^12+2)mod x^97 +x^12+2 i need b(x)=a(x)^3 and i need ans in the form b(0)=a93+a89+a0; b(1)=a65a61; b(2)=a33; . . . b(96)=? plz send the the whole formula AND the complete equations...thanks 
December 14th, 2013, 10:02 PM  #2 
Newbie Joined: Dec 2013 Posts: 3 Thanks: 0  Re: Cube of a polynomial
and one thing more this all calculation is done in F3^97 where F is the field

December 14th, 2013, 10:11 PM  #3 
Newbie Joined: Dec 2013 Posts: 3 Thanks: 0  Re: Cube of a polynomial
Formula to finding cube is b(x)=a(x)^3 =(summation(0 to 96)aix^3i)mod x^97+x^12+2


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