1. **State the problem:** We want to find the value of the constant $c$ such that the piecewise function $$f(x) = \begin{cases} cx^2 + 6x & \text{if } x < 3 \\ x^3 - cx & \text{if } x \geq 3 \end{cases}$$ is continuous on $(-\infty, \infty)$. 2. **Recall the continuity condition:** A function is continuous at $x=3$ if $$\lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x).$$ 3. **Calculate the left-hand limit:** $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (cx^2 + 6x) = c(3)^2 + 6(3) = 9c + 18.$$ 4. **Calculate the right-hand limit and function value at 3:** $$f(3) = 3^3 - c(3) = 27 - 3c.$$ 5. **Set the limits equal for continuity:** $$9c + 18 = 27 - 3c.$$ 6. **Solve for $c$:** $$9c + 3c = 27 - 18$$ $$12c = 9$$ $$c = \frac{9}{12} = \frac{3}{4}.$$ **Final answer:** $$\boxed{c = \frac{3}{4}}.$$