1. **State the problem:** Simplify the expression $$\left(\frac{1}{x} + x\right)^8$$. 2. **Recall the formula:** The expression is a binomial raised to the 8th power. The binomial theorem states: $$\left(a + b\right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ where $a = \frac{1}{x}$, $b = x$, and $n = 8$. 3. **Apply the binomial theorem:** $$\left(\frac{1}{x} + x\right)^8 = \sum_{k=0}^8 \binom{8}{k} \left(\frac{1}{x}\right)^{8-k} x^k$$ 4. **Simplify each term:** $$\binom{8}{k} \frac{x^k}{x^{8-k}} = \binom{8}{k} x^{k - (8-k)} = \binom{8}{k} x^{2k - 8}$$ 5. **Write the full expansion:** $$\sum_{k=0}^8 \binom{8}{k} x^{2k - 8}$$ 6. **Interpretation:** The powers of $x$ go from $-8$ (when $k=0$) to $8$ (when $k=8$) in steps of 2. **Final answer:** $$\left(\frac{1}{x} + x\right)^8 = \sum_{k=0}^8 \binom{8}{k} x^{2k - 8}$$