1. **State the problem:** We have a piecewise function defined as $$f(x) = \begin{cases} k^3 + x & \text{if } x < 3 \\ \frac{16}{k^2 - x} & \text{if } x \geq 3 \end{cases}$$ where $k$ is a positive constant. We want to analyze this function, typically to find $k$ such that $f$ is continuous at $x=3$. 2. **Recall the continuity condition:** For $f$ to be continuous at $x=3$, the left-hand limit and right-hand limit at $x=3$ must be equal to $f(3)$: $$\lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x)$$ 3. **Evaluate the left-hand limit:** Since $x<3$, use the first piece: $$\lim_{x \to 3^-} f(x) = k^3 + 3$$ 4. **Evaluate the right-hand limit and function value at $x=3$:** Since $x \geq 3$, use the second piece: $$f(3) = \frac{16}{k^2 - 3}$$ 5. **Set the limits equal for continuity:** $$k^3 + 3 = \frac{16}{k^2 - 3}$$ 6. **Solve for $k$:** Multiply both sides by $k^2 - 3$: $$\cancel{k^3 + 3} \times \cancel{k^2 - 3} = \cancel{\frac{16}{k^2 - 3}} \times \cancel{k^2 - 3}$$ $$ (k^3 + 3)(k^2 - 3) = 16 $$ 7. **Expand the left side:** $$ k^3 \cdot k^2 - k^3 \cdot 3 + 3 \cdot k^2 - 3 \cdot 3 = 16 $$ $$ k^5 - 3k^3 + 3k^2 - 9 = 16 $$ 8. **Bring all terms to one side:** $$ k^5 - 3k^3 + 3k^2 - 9 - 16 = 0 $$ $$ k^5 - 3k^3 + 3k^2 - 25 = 0 $$ 9. **Interpretation:** This is a quintic equation in $k$. Since $k$ is positive, we look for positive roots. Exact algebraic solutions are complicated, so numerical methods or graphing are recommended. **Final answer:** The value of $k$ satisfies $$ k^5 - 3k^3 + 3k^2 - 25 = 0 $$ with $k > 0$ to ensure continuity of $f$ at $x=3$.