To solve this problem, we need to determine the coefficients $a$, $b$, $c$, and $d$ such that the function $f(x) = ax^3 + bx^2 + cx + d$ has a relative maximum at $(-2, 8)$ and a relative minimum at $(1, -19)$.

Step-by-step solution:

1. Determine critical points:

   • Since $(-2, 8)$ is a relative maximum and $(1, -19)$ is a relative minimum, the derivative $f'(x)$ must be zero at $x = -2$ and $x = 1$.

2. Find $f'(x)$:

   • $f'(x) = 3ax^2 + 2bx + c$.
   • Set $f'(-2) = 0$ and $f'(1) = 0$:
     • At $x = -2$: $3a(-2)^2 + 2b(-2) + c = 0$ $12a - 4b + c = 0$ ... (Equation 1)
     • At $x = 1$: $3a(1)^2 + 2b(1) + c = 0$ $3a + 2b + c = 0$ ... (Equation 2)

3. Use given points:

   • Use the given points $(-2, 8)$ and $(1, -19)$:
     • At $x = -2$, $f(-2) = 8$: $a(-2)^3 + b(-2)^2 + c(-2) + d = 8$ $-8a + 4b - 2c + d = 8$ ... (Equation 3)
     • At $x = 1$, $f(1) = -19$: $a(1)^3 + b(1)^2 + c(1) + d = -19$ $a + b + c + d = -19$ ... (Equation 4)

4. Solve the system of equations: Now, we have four equations:

   • $12a - 4b + c = 0$ (Equation 1)
   • $3a + 2b + c = 0$ (Equation 2)
   • $-8a + 4b - 2c + d = 8$ (Equation 3)
   • $a + b + c + d = -19$ (Equation 4)

   We can solve this system for $a$, $b$, $c$, and $d$.

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