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 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded 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

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