1996 AIME Problem 5

Question

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.

Answer

Use Vieta’s formulas on the original cubic

Expand

(a+b)(b+c)(c+a)

in symmetric sums

Vieta’s formulas give

\begin{aligned}a+b+c&=-3,\\ab+bc+ca&=4,\\abc&=11.\end{aligned}

Also,

\begin{gathered}(a+b)(b+c)(c+a)\\=(a+b+c)(ab+bc+ca)\\\quad-abc=-23.\end{gathered}

This is the product of the roots of the second monic cubic, so its constant term is the negative of that product. Hence

t=23.

Problem 5 in other years