To get a constant term, the exponents of $x$ must cancel.  If we take the term with 2 $x^3$'s and 3 $\frac{1}{x^2}$'s, then they will cancel.  By the Binomial Theorem, this term is $$\binom52 (10x^3)^2\left(-\frac{1}{2x^2}\right)^3=10\cdot100\cdot-\frac{1}{8}\cdot x^6\cdot\frac{1}{x^6}$$$$\Rightarrow \frac{1000}{-8}=\boxed{-125}$$
<|im_start|>system
You are a helpful assistant.<|im_end|>
<|im_start|>user
Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$
Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
<|im_start|>assistant

user
-125
Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$
test