theorem Th18:
  Polynom(1,0,p,q,y) = 0 implies for u,v st y = u+v & 3*v*u+p = 0
  holds u |^ 3 + v |^ 3 = -q & (u |^ 3)*(v |^ 3) = (-p/3) |^ 3
proof
  assume
A1: Polynom(1,0,p,q,y) = 0;
  let u,v;
  assume that
A2: y = u+v and
A3: 3*v*u+p = 0;
  (u+v) |^ 3 = u |^ 3+((3*v)*u^2+(3*v^2)*u)+v |^ 3 by Th14
    .= u |^ 3+(3*v*u)*(u+v)+v |^ 3;
  then 0 = (u |^ 3+v |^ 3)+((3*v*u + p )*(u+v)) + q by A1,A2;
  hence u |^ 3+v |^ 3 = - q by A3;
  3*(v*u) = - p by A3;
  hence thesis by NEWTON:7;
end;
