To solve the equation \((2x-1)^3 = \frac{8}{27}\), we can start by taking the cube root of both sides.

1. Take the cube root of both sides:

\[
2x - 1 = \sqrt[3]{\frac{8}{27}}
\]

2. Simplify the cube root:

\[
\sqrt[3]{\frac{8}{27}} = \frac{\sqrt[3]{8}}{\sqrt[3]{27}} = \frac{2}{3}
\]

So we have:

\[
2x - 1 = \frac{2}{3}
\]

3. Now, solve for \(x\):

Add \(1\) to both sides:

\[
2x = \frac{2}{3} + 1
\]

Convert \(1\) into a fraction with a denominator of \(3\):

\[
1 = \frac{3}{3}
\]

Thus, we have:

\[
2x = \frac{2}{3} + \frac{3}{3} = \frac{5}{3}
\]

4. Finally, divide by \(2\):

\[
x = \frac{5}{3} \cdot \frac{1}{2} = \frac{5}{6}
\]

So, the solution is:

\[
x = \frac{5}{6}
\]