1996 AIME Problems/Problem 5

Problem

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 $t$ .

Solution

By Vieta's formulas on the polynomial $P(x) = x^3+3x^2+4x-11 = (x-a)(x-b)(x-c) = 0$ , we have $a + b + c = s = -3$ , $ab + bc + ca = 4$ , and $abc = 11$ . Then

$t = -(a+b)(b+c)(c+a) = -(s-a)(s-b)(s-c) = -(-3-a)(-3-b)(-3-c)$

This is just the definition for $-P(-3) = \boxed{023}$ .

Alternatively, we can expand the expression to get $\begin{align*} t &= -(-3-a)(-3-b)(-3-c)\\ &= (a+3)(b+3)(c+3)\\ &= abc + 3[ab + bc + ca] + 9[a + b + c] + 27\\ t &= 11 + 3(4) + 9(-3) + 27 = 23\end{align*}$

A third solution arises if it is seen that each term in the expansion of $(a+b)(b+c)(c+a)$ has a total degree of 3. Another way to get terms with degree 3 is to multiply out $(a+b+c)(ab+bc+ca)$ . Expanding both of these expressions and comparing them shows that:

$(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$ $t = -(a+b)(b+c)(c+a) = abc-(ab+bc+ca)(a+b+c) = 11-(4)(-3) = 23$

1996 AIME (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1996 AIME Problems/Problem 5

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