Problem 1406

1. Given that $P(x) = ax^2 + bx + c$ has exactly one real root, we know that the discriminant of $P(x)$ must be zero. The discriminant of a quadratic equation $ax^2 + bx + c$ is given by:
$$\Delta = b^2 - 4ac$$
For $P(x)$ to have exactly one real root, we must have:
$$\Delta = 0 \implies b^2 - 4ac = 0$$

2. Since $P(x)$ has exactly one real root, it can be written in the form:
$$P(x) = a(x - k)^2$$
where $k$ is the root of $P(x)$.

3. We need to find the minimum possible value of $abc$ given that $P(P(P(x)))$ has exactly 3 different real roots. First, let's express $P(P(x))$:
$$P(P(x)) = a(P(x) - k)^2 = a(a(x - k)^2 - k)^2$$

4. Now, we need to find $P(P(P(x)))$:
$$P(P(P(x))) = a(P(P(x)) - k)^2 = a\left(a(a(x - k)^2 - k)^2 - k\right)^2$$

5. For $P(P(P(x)))$ to have exactly 3 different real roots, the polynomial $Q(x) = a\left(a(x - k)^2 - k\right)^2 - k$ must have exactly 3 roots. Since $Q(x)$ is a polynomial of degree 4, one of the roots must be repeated.

6. The repeated root must also be a root of the derivative $Q'(x)$. Let's find $Q'(x)$:
$$Q(x) = a\left(a(x - k)^2 - k\right)^2 - k$$
$$Q'(x) = 4a^2\left(a(x - k)^2 - k\right)(x - k)$$

7. For $Q'(x) = 0$, we have two cases:
$$4a^2\left(a(x - k)^2 - k\right)(x - k) = 0$$
This implies either $x = k$ or $a(x - k)^2 - k = 0$.

8. Since $k \neq 0$, we solve $a(x - k)^2 - k = 0$:
$$a(x - k)^2 = k \implies (x - k)^2 = \frac{k}{a} \implies x - k = \pm \sqrt{\frac{k}{a}}$$
$$x = k \pm \sqrt{\frac{k}{a}}$$

9. Therefore, the roots of $P(P(P(x)))$ are $k$, $k + \sqrt{\frac{k}{a}}$, and $k - \sqrt{\frac{k}{a}}$.

10. To find the minimum possible value of $abc$, we use the fact that $P(x) = a(x - k)^2$. Expanding this, we get:
$$P(x) = ax^2 - 2akx + ak^2$$
Comparing with $P(x) = ax^2 + bx + c$, we have:
$$b = -2ak, \quad c = ak^2$$

11. Using the condition $b^2 - 4ac = 0$:
$$(-2ak)^2 - 4a(ak^2) = 0 \implies 4a^2k^2 - 4a^2k^2 = 0$$
This condition is satisfied.

12. To minimize $abc$:
$$abc = a(-2ak)(ak^2) = -2a^2k^3$$
Using $k = \frac{1}{a}$:
$$abc = -2a^2\left(\frac{1}{a}\right)^3 = -2a^2 \cdot \frac{1}{a^3} = -2 \cdot \frac{1}{a} = -2$$

The final answer is $\boxed{-2}$