1. **State the problem:** Find the limit as $x \to \infty$ of $$\left(\frac{\sqrt[3]{2} + \sqrt[3]{4} + \sqrt[3]{8}}{3}\right)^2$$. 2. **Understand the expression:** The expression inside the parentheses is the average of the cube roots of 2, 4, and 8. 3. **Calculate each cube root:** - $\sqrt[3]{2} = 2^{1/3}$ - $\sqrt[3]{4} = 4^{1/3} = (2^2)^{1/3} = 2^{2/3}$ - $\sqrt[3]{8} = 8^{1/3} = (2^3)^{1/3} = 2^{3/3} = 2$ 4. **Sum the cube roots:** $$2^{1/3} + 2^{2/3} + 2$$ 5. **Divide by 3 to find the average:** $$\frac{2^{1/3} + 2^{2/3} + 2}{3}$$ 6. **Square the average:** $$\left(\frac{2^{1/3} + 2^{2/3} + 2}{3}\right)^2$$ 7. **Since the expression does not depend on $x$, the limit as $x \to \infty$ is just the value of the expression itself.** **Final answer:** $$\boxed{\left(\frac{2^{1/3} + 2^{2/3} + 2}{3}\right)^2}$$