Okay, so I need to find the constant term in the expansion of (10x³ - 1/(2x²))⁵. Hmm, let me think about how to approach this. I remember that when expanding a binomial like (a + b)^n, each term is given by the binomial theorem, which is C(n, k) * a^(n-k) * b^k. So maybe I can apply that here.

First, let's identify a and b in this case. The expression is (10x³ - 1/(2x²))⁵, so a would be 10x³ and b would be -1/(2x²). The exponent n is 5. So each term in the expansion will be C(5, k) * (10x³)^(5 - k) * (-1/(2x²))^k.

Now, the goal is to find the constant term. That means the exponent of x in that term should be zero. So I need to find the value of k where the power of x in the term is zero. Let's break it down.

First, let's write out the general term. The term is C(5, k) * (10x³)^(5 - k) * (-1/(2x²))^k. Let's simplify each part step by step.

Starting with (10x³)^(5 - k). That would be 10^(5 - k) * x^(3*(5 - k)). Then, the second part is (-1/(2x²))^k which is (-1)^k * (1/2)^k * x^(-2k). Multiplying all these together, the coefficients and the powers of x.

So putting it all together, the term becomes:

C(5, k) * 10^(5 - k) * (-1)^k * (1/2)^k * x^(3*(5 - k) - 2k).

Simplify the exponent of x:

3*(5 - k) - 2k = 15 - 3k - 2k = 15 - 5k.

We need 15 - 5k = 0 for the term to be constant. Solving for k:

15 - 5k = 0 → 5k = 15 → k = 3.

So when k = 3, the exponent of x is zero, which means that term is the constant term. Now, let's calculate that term.

First, compute the binomial coefficient C(5, 3). That's 5! / (3! * (5 - 3)!) = (120)/(6 * 2) = 120 / 12 = 10.

Next, compute 10^(5 - 3) = 10² = 100.

Then, (-1)^3 = -1.

(1/2)^3 = 1/8.

Multiply all these together with the binomial coefficient:

10 * 100 * (-1) * (1/8) = 10 * 100 * (-1/8) = 10 * (-12.5) = -125.

Wait, let me check that again. 10 * 100 is 1000, then 1000 * (-1) is -1000, then -1000 * (1/8) is -125. Yeah, that's right. So the constant term is -125.

But let me double-check my steps to make sure I didn't make a mistake. Let's verify each part.

First, binomial coefficient C(5, 3). Yes, 5 choose 3 is 10. Correct.

Then, 10^(5 - 3) = 10² = 100. Correct.

(-1)^3 = -1. Correct.

(1/2)^3 = 1/8. Correct.

Multiply all together: 10 * 100 = 1000. 1000 * (-1) = -1000. -1000 * (1/8) = -125. Yep, that seems right.

Alternatively, maybe I can check by expanding the entire expression, but that would be tedious. Alternatively, maybe use another method to verify.

Alternatively, consider the exponents. The exponent of x in each term is 15 - 5k. We set that to zero, so k=3. Then, the term is C(5,3)*(10x³)^(2)*(-1/(2x²))^3. Let's compute that again.

(10x³)^2 = 100x^6.

(-1/(2x²))^3 = (-1)^3/(2^3 x^(6)) = -1/8 x^(-6).

Multiplying these together: 100x^6 * (-1/8 x^-6) = 100*(-1/8) x^(6-6) = -12.5 x^0 = -12.5.

Then multiply by the binomial coefficient 10: 10 * (-12.5) = -125. Yes, same result.

So that seems correct. Therefore, the constant term is -125.

I think that's solid. Let me just recap to ensure I didn't skip any steps or make any miscalculations.

1. Identify a and b in the binomial: a = 10x³, b = -1/(2x²). Exponent n = 5.

2. General term: C(5,k) * (10x³)^(5 - k) * (-1/(2x²))^k.

3. Simplify the exponents: 3*(5 - k) - 2k = 15 - 5k. Set to zero: k=3.

4. Compute the term for k=3:

   - Binomial coefficient: 10.

   - (10x³)^2 = 100x^6.

   - (-1/(2x²))^3 = -1/8 x^-6.

   - Multiply together: 100*(-1/8) x^0 = -12.5.

   - Multiply by 10: -125.

All steps check out. Therefore, the constant term is -125.

**Final Answer**
The constant term is \boxed{-125}.
Therefore, the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$ is $\boxed{-125}$.
</think>