Divide 184 into two parts such that one-third of one part may exceed one-seventh of the other part by 8.

Let the first part be x.

Then, the second part = (184 - x).

According to the question, one-third of the first part exceeds one-seventh of the second part by 8:

$\therefore \dfrac{1}{3}x - \dfrac{1}{7}(184 - x) = 8 \\[1em] \Rightarrow \dfrac{7x - 3(184 - x)}{21} = 8 \\[1em] \Rightarrow 7x - 552 + 3x = 8 \times 21 \\[1em] \Rightarrow 10x - 552 = 168 \\[1em] \Rightarrow 10x = 168 + 552 \quad \text{[Transposing -552 to RHS]} \\[1em] \Rightarrow 10x = 720 \\[1em] \Rightarrow x = \dfrac{720}{10} \\[1em] \Rightarrow x = 72$

First part = x = 72.

Second part = (184 - x) = (184 - 72) = 112.

Hence, the two parts are 72 and 112.