To solve the equation (2x - 1)^2 = 125, we can take the square root of both sides:

√((2x - 1)^2) = √125

2x - 1 = ±√125

2x - 1 = ±5√5

Now we can solve for x:

Case 1: 2x - 1 = 5√5
2x = 1 + 5√5
x = (1 + 5√5)/2

Case 2: 2x - 1 = -5√5
2x = 1 - 5√5
x = (1 - 5√5)/2

Therefore, the solutions to the equation (2x - 1)^2 = 125 are x = (1 + 5√5)/2 and x = (1 - 5√5)/2.

To solve the equation 4x^3 + 15 = 19, we can subtract 15 from both sides:

4x^3 = 19 - 15
4x^3 = 4

Now we can solve for x by taking the cube root of both sides:

∛(4x^3) = ∛4

x = ∛4

Therefore, the solution to the equation 4x^3 + 15 = 19 is x = ∛4.

To solve the equation (4x - 1)^2 = 25 * 9, we can simplify the right side first:

(4x - 1)^2 = 225

Now we can take the square root of both sides:

√((4x - 1)^2) = √225

4x - 1 = ±15

Now we can solve for x:

Case 1: 4x - 1 = 15
4x = 1 + 15
x = 16/4
x = 4

Case 2: 4x - 1 = -15
4x = 1 - 15
x = -14/4
x = -7/2

Therefore, the solutions to the equation (4x - 1)^2 = 25 * 9 are x = 4 and x = -7/2.