1. **State the problem:** We have the function $$f(x) = \frac{8x^3 - 27}{2x - 3}$$ which has a removable discontinuity at $$x = \frac{3}{2}$$. We want to find which function definition removes this discontinuity. 2. **Recall the concept:** A removable discontinuity occurs when the function is undefined at a point but the limit exists. To remove it, we redefine the function at that point to equal the limit. 3. **Factor the numerator:** Recognize that $$8x^3 - 27$$ is a difference of cubes: $$8x^3 - 27 = (2x)^3 - 3^3 = (2x - 3)(4x^2 + 6x + 9)$$ 4. **Simplify the function:** $$f(x) = \frac{(2x - 3)(4x^2 + 6x + 9)}{2x - 3}$$ For $$x \neq \frac{3}{2}$$, the $$2x - 3$$ terms cancel: $$f(x) = 4x^2 + 6x + 9$$ 5. **Find the limit at the discontinuity:** $$\lim_{x \to \frac{3}{2}} f(x) = 4\left(\frac{3}{2}\right)^2 + 6\left(\frac{3}{2}\right) + 9$$ Calculate step-by-step: $$4 \times \frac{9}{4} + 6 \times \frac{3}{2} + 9 = 9 + 9 + 9 = 27$$ 6. **Conclusion:** To remove the discontinuity, define $$f\left(\frac{3}{2}\right) = 27$$. **Answer:** The function that removes the discontinuity is: $$f(x) = \begin{cases} \frac{8x^3 - 27}{2x - 3}, & x \neq \frac{3}{2} \\ 27, & x = \frac{3}{2} \end{cases}$$