Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta's formula.
Suppose that the roots of \(x^{3}+3x^{2}+4x-11=0\) are a,b and c and that the roots of \(x^{3}+rx^{2}+sx+t=0\) are a+b,b+c and c+a, find t.
Functions
Roots of Equation
Vieta s formula
Answer: is 23.
AIME I, 1996, Question 5
Polynomials by Barbeau
With Vieta s formula
\(f(x)=x^{3}+3x^{2}+4x-11=(x-a)(x-b)(x-c)=0\)
\(\Rightarrow a+b+c=-3\), \(ab+bc+ca=4\) and \(abc=11\)
Let a+b+c=-3=p
here t=-(a+b)(b+c)(c+a)
\(\Rightarrow t=-(p-c)(p-a)(p-b)\)
\(\Rightarrow t=-f(p)=-f(-3)\)
\(t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]\)
=23.
