Source: 1976 Euclid Part A Problem 8
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Given that $a$, $b$, and $c$ are the roots of the equation $x^3-3x^2+mx+24=0$, and that $-a$ and $-b$ are the roots of the equation $x^2+nx-6=0$, then the value of $n$ is

$\textbf{(A) } 1 \qquad \textbf{(B) } -1 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } -7 \qquad \textbf{(E) } \text{none of these}$
1. Given the polynomial \(x^3 - 3x^2 + mx + 24 = 0\) with roots \(a\), \(b\), and \(c\), we can use Vieta's formulas to find the relationships between the coefficients and the roots:
   \[
   a + b + c = 3,
   \]
   \[
   ab + bc + ca = m,
   \]
   \[
   abc = -24.
   \]

2. Given the polynomial \(x^2 + nx - 6 = 0\) with roots \(-a\) and \(-b\), we can use Vieta's formulas again:
   \[
   (-a) + (-b) = -n \implies -(a + b) = -n \implies a + b = n,
   \]
   \[
   (-a)(-b) = ab = -6.
   \]

3. From the equation \(ab = -6\) and \(abc = -24\), we can solve for \(c\):
   \[
   c = \frac{abc}{ab} = \frac{-24}{-6} = 4.
   \]

4. Substitute \(c = 4\) into the equation \(a + b + c = 3\):
   \[
   a + b + 4 = 3 \implies a + b = 3 - 4 = -1.
   \]

5. Since \(a + b = n\), we have:
   \[
   n = -1.
   \]

The final answer is \(\boxed{-1}\).