To solve the equation ∛(x-2) + √(2x-1) = 3, we can start by isolating one of the radicals. Let's isolate the cube root (∛) term.

∛(x-2) = 3 - √(2x-1)

Now, cube both sides of the equation to eliminate the cube root:

(∛(x-2))^3 = (3 - √(2x-1))^3

Simplifying the left side:

x - 2 = (3 - √(2x-1))^3

Expand the right side using the binomial formula:

x - 2 = (3 - √(2x-1))(3 - √(2x-1))(3 - √(2x-1))

Multiply the first two terms:

x - 2 = (9 - 3√(2x-1) - 3√(2x-1) + (2x-1))

Simplify:

x - 2 = (9 - 6√(2x-1) + 2x - 1)

Combine like terms:

x - 2 = (8 + 2x - 6√(2x-1))

Rearrange the equation:

x - 2 - 2x = 8 - 6√(2x-1)

Simplify:

-x - 2 = 8 - 6√(2x-1)

Add x to both sides:

-2 = x + 8 - 6√(2x-1)

Subtract 8 from both sides:

-10 = x - 6√(2x-1)

Add 6√(2x-1) to both sides:

6√(2x-1) - 10 = x

Now, let's isolate the square root (√) term:

√(2x-1) = x - 6√(2x-1) - 10

Add 6√(2x-1) to both sides:

7√(2x-1) = x - 10

Square both sides to eliminate the square root:

(7√(2x-1))^2 = (x - 10)^2

Simplifying the left side:

49(2x-1) = (x - 10)^2

Expand the right side:

49(2x-1) = x^2 - 20x + 100

Distribute on the left side:

98x - 49 = x^2 - 20x + 100

Rearrange the equation:

x^2 - 118x + 149 = 0

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. However, upon inspection, it seems that this quadratic equation does not have any rational solutions. Therefore, we can conclude that there are no real solutions to the original equation ∛(x-2) + √(2x-1) = 3.