Question 1209706
Let $P(x) = x^2 - x - 3$. The roots of $P(x)$ are given by the quadratic formula:
$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{1+12}}{2} = \frac{1 \pm \sqrt{13}}{2}$$
Let $r_1 = \frac{1 + \sqrt{13}}{2}$ and $r_2 = \frac{1 - \sqrt{13}}{2}$.
Since $r_1$ and $r_2$ are roots of $x^2 - x - 3 = 0$, we have
$r_1^2 - r_1 - 3 = 0$ and $r_2^2 - r_2 - 3 = 0$.
Thus, $r_1^2 = r_1 + 3$ and $r_2^2 = r_2 + 3$.
Also, $r_1 + r_2 = 1$ and $r_1 r_2 = -3$.

Let $Q(x) = x^3 - bx^2 - c$. Since $r_1$ and $r_2$ are roots of $Q(x)$, we have
$r_1^3 - br_1^2 - c = 0$ and $r_2^3 - br_2^2 - c = 0$.
$r_1^3 = r_1 \cdot r_1^2 = r_1(r_1+3) = r_1^2 + 3r_1 = (r_1+3) + 3r_1 = 4r_1 + 3$
$r_2^3 = r_2 \cdot r_2^2 = r_2(r_2+3) = r_2^2 + 3r_2 = (r_2+3) + 3r_2 = 4r_2 + 3$
Substituting these into the equations for $Q(x)$, we get
$4r_1 + 3 - b(r_1 + 3) - c = 0$
$4r_2 + 3 - b(r_2 + 3) - c = 0$
Subtracting the second equation from the first, we get
$4(r_1 - r_2) - b(r_1 - r_2) = 0$
$r_1 - r_2 \ne 0$, so $4 - b = 0$, which gives $b = 4$.
Substituting $b=4$ into the first equation, we get
$4r_1 + 3 - 4(r_1 + 3) - c = 0$
$4r_1 + 3 - 4r_1 - 12 - c = 0$
$-9 - c = 0$
$c = -9$
Thus, $(b, c) = (4, -9)$.

Final Answer: The final answer is $\boxed{(4,-9)}$