Find the term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ which does not contain $x$.
By binomial theorem, $\;$ $\left(a + b\right)^n = C^n_0 \; a^n + C^n_1 \; a^{n-1} \; b + \cdots + C^n_k \; a^{n-k} \; b^k + \cdots + C^n_n \; b^n$
$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$
Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ be independent of $x$.
Now, $\;$ $T_{k+1} = C^8_k \; x^{8-k} \; \left(\dfrac{1}{x}\right)^{k} = C^8_k \; x^{8-k-k} = C^8_k \; x^{8-2k}$
Since this term is independent of $x$ $\implies$ $8 - 2k = 0$ $\implies$ $k = 4$
i.e. $5^{th}$ term is independent of $x$.
$\therefore \;$ The term independent of $x$ in the expansion of $\left(x + \dfrac{1}{x}\right)^8$ is
$T_5 = T_{4+1} = C^8_4 = \dfrac{8!}{4! \left(8 - 4\right)!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4!}{4! \times 4 \times 3 \times 2 \times 1} = 70$