1. **State the problem:** We want to find the value of $k$ such that the piecewise function $$f(x) = \begin{cases} k^3 + x & \text{if } x < 3 \\ \frac{16}{k^2 - x} & \text{if } x \geq 3 \end{cases}$$ is continuous at $x=3$. 2. **Continuity condition:** For $f$ to be continuous at $x=3$, the left-hand limit and right-hand limit at $x=3$ must be equal, and equal to $f(3)$. This means: $$\lim_{x \to 3^-} f(x) = f(3) = \lim_{x \to 3^+} f(x)$$ 3. **Evaluate left-hand limit:** For $x<3$, $$f(x) = k^3 + x$$ So, $$\lim_{x \to 3^-} f(x) = k^3 + 3$$ 4. **Evaluate right-hand limit and function value at $x=3$:** For $x \geq 3$, $$f(x) = \frac{16}{k^2 - x}$$ So, $$f(3) = \frac{16}{k^2 - 3}$$ 5. **Set continuity condition:** $$k^3 + 3 = \frac{16}{k^2 - 3}$$ 6. **Solve for $k$:** Multiply both sides by $k^2 - 3$: $$ (k^3 + 3)(k^2 - 3) = 16 $$ 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. **Solve the equation:** We look for positive roots of $$ k^5 - 3k^3 + 3k^2 - 25 = 0 $$ 10. **Check given options:** - For $k=2.081$, approximate left side: $$2.081^5 - 3(2.081)^3 + 3(2.081)^2 - 25 \approx 0$$ (close to zero) - For $k=2.646$ and $k=8.550$, values do not satisfy the equation. 11. **Conclusion:** The value $k \approx 2.081$ makes $f$ continuous at $x=3$. **Final answer:** $\boxed{2.081}$