To solve the equation \((2x-1)^3 + (x-2)^3 + (3-3x)^3 = 0\), we can use the identity for the sum of three cubes:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \]

First, let's identify \(a\), \(b\), and \(c\) in our equation:

\[ a = 2x - 1 \]
\[ b = x - 2 \]
\[ c = 3 - 3x \]

We need to check if \(a + b + c = 0\):

\[ (2x - 1) + (x - 2) + (3 - 3x) \]
\[ = 2x - 1 + x - 2 + 3 - 3x \]
\[ = 2x + x - 3x - 1 - 2 + 3 \]
\[ = 0 \]

Since \(a + b + c = 0\), the identity simplifies to:

\[ a^3 + b^3 + c^3 = 3abc \]

Thus, our equation becomes:

\[ (2x - 1)^3 + (x - 2)^3 + (3 - 3x)^3 = 3(2x - 1)(x - 2)(3 - 3x) \]

Given that the left-hand side is zero, we have:

\[ 0 = 3(2x - 1)(x - 2)(3 - 3x) \]

This implies that at least one of the factors on the right-hand side must be zero:

1. \(2x - 1 = 0\)
2. \(x - 2 = 0\)
3. \(3 - 3x = 0\)

Solving these equations:

1. \(2x - 1 = 0\)
\[ 2x = 1 \]
\[ x = \frac{1}{2} \]

2. \(x - 2 = 0\)
\[ x = 2 \]

3. \(3 - 3x = 0\)
\[ 3 = 3x \]
\[ x = 1 \]

Therefore, the solutions to the equation \((2x-1)^3 + (x-2)^3 + (3-3x)^3 = 0\) are:

\[ x = \frac{1}{2}, \; x = 2, \; x = 1 \]