Suppose that the [roots](https://artofproblemsolving.com/wiki/index.php/Root) 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 $t$.
By [Vieta's formulas](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_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*}