# 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 $t$.

~ pi_is_3.14

## Solution 1

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*}

## Solution 2

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:

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

A way to realize $(a+b)(b+c)(c+a) = (ab+bc+ca)(a+b+c)-abc$:

$(a + b)(b+c)(c+a) = a^2b + a^2c + b^2a + b^2c + c^2a + c^2b + abc + abc$

(Add an extra $abc$)

$a^2b + a^2c + abc + b^2a + b^2c + abc + c^2a + c^2b + abc - abc =$

$a(ab + bc + ac) + b(ab + bc + ac) + c(ab + bc + ac) - abc =$

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

The value of $t$ is the negation of this, which is $-(-23) = \boxed{023}$

~Extremelysupercooldude

## Solution 3

We have that $x^3+3x^2+4x-11=0$ for roots $a,b,c.$ In the second cubic function $x^3+rx^2+sx+t=0,$ the roots are $a+b,b+c,c+a.$

By Vieta's formulae, we see that $t= -(a+b)(b+c)(a+c).$ As we know that the sum of the roots of the first polynomial, $a+b+c$ is $-3$ by applying Vieta's again.

Using this fact, we can rewrite $t$ as $-(-3-a)(-3-b)(-3-c) = (a+3)(b+3)(c+3).$

Seeing this, we can find the value of the product of the roots by applying this to the first equation. This can be done by setting $x'=a+3,$ so, from this, we see that we should plug in $x'-3$ for $x$ in $x^3+3x^2+4x-11=0$.

After simplifying, we get that the polynomial is $x'^3 - 6x'^2 + 13x' - 23 = 0.$ Given that the product of the roots of this equation is equivalent to our desired value for $t$, we can apply Vieta's formulae for a third time to find that $t = -(\frac{-23}{1}) = \boxed{023}.$

~MathWhiz35