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December 14th, 2013, 09:51 PM   #1
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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)=a65-a61;
b(2)=a33;
.
.
.
b(96)=?
plz send the the whole formula AND the complete equations...thanks
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December 14th, 2013, 10:02 PM   #2
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Re: Cube of a polynomial

and one thing more this all calculation is done in F3^97 where F is the field
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December 14th, 2013, 10:11 PM   #3
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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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