Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Roots of Equation and Vieta’s formula.

## Roots of Equation and Vieta’s formula – AIME I, 1996

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.

- is 107
- is 23
- is 840
- cannot be determined from the given information

**Key Concepts**

Functions

Roots of Equation

Vieta s formula

## Check the Answer

But try the problem first…

Answer: is 23.

AIME I, 1996, Question 5

Polynomials by Barbeau

## Try with Hints

First hint

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 \(abc=11\)

Second Hint

Let a+b+c=-3=p

here t=-(a+b)(b+c)(c+a)

\(\Rightarrow t=-(p-c)(p-a)(p-b)\)

\(\Rightarrow t=-f(p)=-f(-3)\)

Final Step

\(t=-[(-3)^{3}+3(-3)^{2}+4(-3)-11]\)

=23.

## Other useful links

- https://www.cheenta.com/cubes-and-rectangles-math-olympiad-hanoi-2018/
- https://www.youtube.com/watch?v=ST58GTF95t4&t=140s

Google