a) To solve the equation x^3 - 25x = 0, we can factor out an x:

x(x^2 - 25) = 0

Now, we have two factors: x = 0 and x^2 - 25 = 0.

For x = 0, we have one solution.

For x^2 - 25 = 0, we can factor it as a difference of squares:

(x - 5)(x + 5) = 0

Now, we have two more solutions: x = 5 and x = -5.

Therefore, the solutions to the equation x^3 - 25x = 0 are x = 0, x = 5, and x = -5.

b) To solve the equation x^3 - 4x^2 - 9x + 36 = 0, we can try factoring it using synthetic division or the rational root theorem. However, in this case, there are no rational roots.

Using a graphing calculator or software, we can find that the equation has one real root at approximately x = 3. This means that (x - 3) is a factor of the equation.

Using polynomial long division or synthetic division, we can divide x^3 - 4x^2 - 9x + 36 by (x - 3):

(x^3 - 4x^2 - 9x + 36) / (x - 3) = x^2 - x - 12

Now, we have factored the equation as (x - 3)(x^2 - x - 12) = 0.

To solve x^2 - x - 12 = 0, we can factor it:

(x - 4)(x + 3) = 0

Now, we have two more solutions: x = 4 and x = -3.

Therefore, the solutions to the equation x^3 - 4x^2 - 9x + 36 = 0 are x = 3, x = 4, and x = -3.