1. **State the problem:** We want to find which term in the expansion of $$\left(\frac{1}{x} + x\right)^8$$ equals 70. 2. **Recall the binomial expansion formula:** $$\left(a + b\right)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$$ 3. **Apply the formula to our problem:** Here, $$a = \frac{1}{x} = x^{-1}$$ and $$b = x$$, with $$n=8$$. The general term $$T_{k+1}$$ is: $$T_{k+1} = \binom{8}{k} \left(x^{-1}\right)^{8-k} \left(x\right)^k = \binom{8}{k} x^{-(8-k)} x^k = \binom{8}{k} x^{k-(8-k)} = \binom{8}{k} x^{2k-8}$$ 4. **Find the term that is a constant (no $$x$$):** For the term to be independent of $$x$$, the exponent must be zero: $$2k - 8 = 0$$ $$2k = 8$$ $$k = 4$$ 5. **Calculate the coefficient of the constant term:** $$T_{5} = \binom{8}{4} x^{0} = \binom{8}{4} = \frac{8!}{4!4!} = \frac{40320}{24 \times 24} = 70$$ 6. **Conclusion:** The 5th term in the expansion is 70 and is the constant term. **Final answer:** The term number is **5** and its value is **70**.