1. **State the problem:** Given $x = 3 + \sqrt{8}$, find the value of $$x^4 + \frac{a}{x^4}$$. However, the problem does not specify the value of $a$. Assuming $a=1$ for the expression to be meaningful and solvable. 2. **Rewrite the problem:** Find $$x^4 + \frac{1}{x^4}$$ where $x = 3 + \sqrt{8}$. 3. **Important formula:** For any $x$, $$x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2$$. 4. **Calculate $x + \frac{1}{x}$:** $$x = 3 + \sqrt{8}$$ Calculate $$\frac{1}{x} = \frac{1}{3 + \sqrt{8}}$$. Rationalize the denominator: $$\frac{1}{3 + \sqrt{8}} \times \frac{3 - \sqrt{8}}{3 - \sqrt{8}} = \frac{3 - \sqrt{8}}{9 - 8} = 3 - \sqrt{8}$$ So, $$x + \frac{1}{x} = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 6$$ 5. **Calculate $x^2 + \frac{1}{x^2}$:** Use the identity: $$\left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}$$ So, $$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = 6^2 - 2 = 36 - 2 = 34$$ 6. **Calculate $x^4 + \frac{1}{x^4}$:** Using the formula from step 3: $$x^4 + \frac{1}{x^4} = \left(x^2 + \frac{1}{x^2}\right)^2 - 2 = 34^2 - 2 = 1156 - 2 = 1154$$ **Final answer:** $$x^4 + \frac{1}{x^4} = 1154$$