Recent content by myro111

Dont worry I solved it another way.But thanks for the try.I used Vietes formulas

1. Divide the second polynomial with x-ab and I get as a remainder (ab)^6+(ab)^4+(ab)^3-(ab)^2-1 The remainder is the second polynomial but instead of x it is ab.I got this idea because if we have a polynomial P(x),a is the root only and only if P(x) is divisible by x-a,and that fact is well...

I found some problems from old contests in my book and I need some help solving them since I could not find the solutions online. Some USA contest problems: 1.(1977.)Let a and b be 2 solutions of x^4+x^3-1=0 .Prove that a*b is the solution of x^6+x^4+x^3-x^2-1=0 2.(1983.).Prove that all...