$8^0 + 8^{-1} + 4^{-1}$ is equal to:

1. $8\dfrac{3}{8}$

2. $1\dfrac{3}{8}$

3. $\dfrac{3}{8}$

4. $2\dfrac{2}{3}$

As we know, for any non-zero rational number a

$a^{-n} = \dfrac{1}{a^n}$ and $a^{n} = \dfrac{1}{a^{-n}}$.

And $a^0 = 1$

Hence,

$8^0 + 8^{-1} + 4^{-1}\\[1em] = 1 + \dfrac{1}{8} + \dfrac{1}{4} \\[1em] = \dfrac{1}{1} + \dfrac{1}{8} + \dfrac{1}{4}$

LCM of 1, 8 and 4 is 2 x 2 x 2 = 8

$= \dfrac{1 \times 8}{1 \times 8} + \dfrac{1 \times 1}{8 \times 1} + \dfrac{1 \times 2}{4 \times 2}\\[1em] = \dfrac{8}{8} + \dfrac{1}{8} + \dfrac{2}{8}\\[1em] = \dfrac{8 + 1 + 2}{8}\\[1em] = \dfrac{11}{8}\\[1em] = 1\dfrac{3}{8}$

Hence, option 2 is the correct option.