Okay, let's tackle this problem step by step. The question is asking for the sum of the inverse function values at 0 and 6, that is, f⁻¹(0) + f⁻¹(6). Since the function f is piecewise-defined, we need to check both pieces for each required inverse value. Let me break this down into smaller tasks.

First, I need to find f⁻¹(0). That means I have to solve f(x) = 0. Since f is defined differently for x ≤ 3 and x > 3, I need to check both cases.

Starting with the first piece: 3 - x. If 3 - x = 0, then x = 3. But wait, x = 3 is the boundary point between the two pieces. Let me check if this is valid. Since the first piece is defined for x ≤ 3, x = 3 is included here. So, f(3) = 3 - 3 = 0. Therefore, f⁻¹(0) is 3.

But wait, maybe there's another solution in the second piece? Let's check the second piece: -x³ + 2x² + 3x. Set this equal to 0 and solve for x > 3.

So, -x³ + 2x² + 3x = 0. Factoring out an x gives x(-x² + 2x + 3) = 0. The solutions are x = 0 and the solutions to -x² + 2x + 3 = 0. Solving the quadratic equation:

-x² + 2x + 3 = 0 => x² - 2x - 3 = 0. Using quadratic formula: x = [2 ± sqrt(4 + 12)] / 2 = [2 ± sqrt(16)] / 2 = [2 ± 4]/2. So x = 3 or x = -1.

But since we are considering x > 3 for the second piece, none of these solutions (0, 3, -1) are in that interval. Therefore, the only solution for f(x) = 0 is x = 3. So, f⁻¹(0) = 3.

Now, moving on to f⁻¹(6). Again, we need to check both pieces.

First, check the first piece: 3 - x = 6. Solving for x gives x = 3 - 6 = -3. But x = -3 is less than or equal to 3, so we need to check if this is valid. Let's compute f(-3): 3 - (-3) = 6. Yes, that works. So, f(-3) = 6, which means f⁻¹(6) = -3.

But again, let's check the second piece just in case there's another solution. Set -x³ + 2x² + 3x = 6. So, -x³ + 2x² + 3x - 6 = 0. Let's try to factor this. Maybe rational roots? The possible rational roots are ±1, ±2, ±3, ±6.

Testing x=1: -1 + 2 + 3 -6 = -2 ≠ 0

x=2: -8 + 8 + 6 -6 = 0. Oh, x=2 is a root. Then, we can factor out (x - 2). Let's perform polynomial division.

Divide -x³ + 2x² + 3x -6 by (x - 2). Using synthetic division:

Coefficients: -1, 2, 3, -6

Bring down -1. Multiply by 2: -2. Add to next coefficient: 2 + (-2) = 0. Multiply by 2: 0. Add to next coefficient: 3 + 0 = 3. Multiply by 2: 6. Add to last coefficient: -6 + 6 = 0. So, the polynomial factors as (x - 2)(-x² + 0x + 3) = 0. Therefore, the roots are x = 2 and the roots of -x² + 3 = 0 => x² = 3 => x = sqrt(3) ≈ 1.732 or x = -sqrt(3) ≈ -1.732.

But in the second piece, x > 3. So, none of these roots (2, sqrt(3), -sqrt(3)) are greater than 3. Therefore, there is no solution in the second piece. So, the only solution is x = -3 in the first piece.

Therefore, f⁻¹(6) is -3.

Adding them together: 3 + (-3) = 0.

Wait, that seems too straightforward. Let me verify once again to make sure I didn't make a mistake.

For f⁻¹(0): solved 3 - x = 0 gives x = 3, which is valid since x=3 is included in the first piece. The second piece equation didn't yield any solutions in x > 3. So, f⁻¹(0) is indeed 3.

For f⁻¹(6): solved 3 - x = 6 gives x = -3, which is valid as x=-3 ≤ 3. The second piece's equation didn't give any solutions in x > 3. So, f⁻¹(6) is indeed -3.

Adding them: 3 + (-3) = 0. So the final answer is 0.

But wait, is there a possibility that the inverse functions could have different branches? For example, maybe the function is not injective? Let me check the function's behavior.

The first piece is a linear function decreasing from x ≤ 3: as x decreases, f(x) increases. So from negative infinity up to x=3, f(x) goes from 3 - (-∞) = ∞ down to 0. So the first piece maps (-∞, 3] to [0, ∞).

The second piece is a cubic function: -x³ + 2x² + 3x for x > 3. Let's analyze its behavior. As x approaches 3 from the right, f(3+) = -27 + 18 + 9 = 0. Wait, but f(3) is 0 from the first piece. So at x=3, both pieces meet at 0. Now, as x increases beyond 3, let's see the behavior of the cubic.

The leading term is -x³, so as x approaches infinity, f(x) approaches -infinity. The derivative of the second piece: f'(x) = -3x² + 4x + 3. To find critical points, set derivative to zero: -3x² + 4x + 3 = 0 => 3x² -4x -3 = 0. Solutions: x = [4 ± sqrt(16 + 36)] / 6 = [4 ± sqrt(52)] /6 = [4 ± 2*sqrt(13)] /6 = [2 ± sqrt(13)] /3. Approximately, sqrt(13) ≈ 3.606, so the critical points are at (2 + 3.606)/3 ≈ 1.869 and (2 - 3.606)/3 ≈ -0.535. So, for x > 3, the derivative is negative (since the leading coefficient is -3). Therefore, the function is decreasing for x > 3. So, as x increases beyond 3, f(x) decreases from 0 to -infinity. Therefore, the second piece maps (3, ∞) to (-∞, 0).

Therefore, the entire function f is bijective? Wait, the first piece maps (-∞, 3] to [0, ∞), and the second piece maps (3, ∞) to (-∞, 0). So, the range of f is (-∞, ∞), but with a jump at x=3. However, since f(3) is 0, but the second piece approaches 0 from the negative side as x approaches 3 from the right. Wait, actually, as x approaches 3 from the right, f(x) approaches -27 + 18 + 9 = 0. So, the second piece starts at 0 (from the right) and decreases. Wait, but the first piece at x=3 is 0. So, the function f is continuous at x=3? Let's check:

Left-hand limit as x approaches 3 from the left: 3 - 3 = 0.

Right-hand limit as x approaches 3 from the right: -27 + 18 + 9 = 0.

So, the function is continuous at x=3. Therefore, the function f is continuous everywhere. Moreover, the first piece is decreasing on (-∞, 3], and the second piece is decreasing on (3, ∞). Therefore, the entire function is decreasing on its entire domain, which is (-∞, ∞). Therefore, f is strictly decreasing everywhere, which means it is bijective (both injective and surjective). Therefore, the inverse function exists and is also strictly decreasing.

Given that f is strictly decreasing, each value in its range corresponds to exactly one x in its domain. Therefore, for each y in (-∞, ∞), there is exactly one x such that f(x) = y. Therefore, the inverse function is well-defined.

Therefore, the previous calculations are correct. f⁻¹(0) = 3 and f⁻¹(6) = -3. Adding them gives 0. So, the answer is 0.

But just to be absolutely thorough, let me check the second inverse value again. Suppose I made a mistake in solving the second equation. Let's re-examine f⁻¹(6). For the first piece, 3 - x = 6, so x = -3. As x=-3 is ≤3, that's valid. Then, check the second piece: solving -x³ + 2x² + 3x = 6. We found that x=2 is a root, but 2 is not in x>3. The other roots are x= sqrt(3) (~1.732) and x=-sqrt(3) (~-1.732), neither of which are greater than 3. Therefore, no solutions in the second piece. Therefore, the only solution is x=-3. So, f⁻¹(6) is indeed -3.

Therefore, the sum is 3 + (-3) = 0. So, the answer is 0. Let me just confirm once more with a different perspective. Since the function is strictly decreasing, the inverse function will also be strictly decreasing. Therefore, if f(a) = 0 and f(b) = 6, then a must be less than b because as x increases, f(x) decreases. Wait, but in this case, a = 3 and b = -3. But -3 < 3, which is correct because f(-3) = 6 and f(3) = 0, so when x increases from -3 to 3, f(x) decreases from 6 to 0. Then, when x increases beyond 3, f(x) decreases further. Therefore, the inverse function at 6 is -3 and at 0 is 3. So, adding them gives 0. That seems consistent.

Alternatively, if I think of the function as a strictly decreasing function, then the inverse function will map 6 to the x that gives 6, which is -3, and 0 to the x that gives 0, which is 3. Since the function is symmetric in a way around x=3, but not exactly. Wait, but the function isn't symmetric. However, the two points f(-3)=6 and f(3)=0 are symmetric with respect to the line x=0. Hmm, not exactly. But their sum is 0. That's interesting. So, even though the function isn't symmetric, the particular points where f(x) = 6 and f(x)=0 happen to be at x=-3 and x=3, which are negatives of each other. Therefore, their sum is zero. That's a nice observation.

So, after all these checks, I believe the answer is indeed 0. I think I've covered all possible cases and verified the calculations thoroughly. There's no mistake in the reasoning here. Therefore, the final answer is 0.
Therefore, the sum of the inverse function values at 0 and 6 is $\boxed{0}$.
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