My Math Forum Cube of a polynomial

 Abstract Algebra Abstract Algebra Math Forum

 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)=a65-a61; 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

 Tags cube, polynomial

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post tejolson Algebra 0 November 4th, 2013 10:13 PM condemath Algebra 3 September 20th, 2011 08:34 PM rayan Kalak Algebra 1 December 28th, 2010 04:58 AM yano Algebra 8 March 25th, 2010 05:40 PM yano Applied Math 4 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top