Question 1207665
.


        You factored incorrectly; making errors, you created difficulties for yourself.


                A correct factoring is below.



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(x^2-3x)^(1/3) * [x + 2*(x^2-3x)^(3/3)) = 0,

or

(x^2-3x)^(1/3) * (2x^2-5x) = 0.


It deploys in separate equations


    (a)  x = 0

    (b)  x - 3 = 0

    (c)  2x - 5 = 0


with elementary solutions.
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