SOLUTION: Find the real solutions by factoring. x(x^2 - 3x)^(1/3) + 2(x^2 - 3x)^(4/3) = 0 I factored out (x^2 - 3x)^(1/3). (x^2 - 3x)^(1/3)[x + 2(x^2 - 3x)^(4/9)] Stuck here....

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Question 1207665: Find the real solutions by factoring.
x(x^2 - 3x)^(1/3) + 2(x^2 - 3x)^(4/3) = 0
I factored out (x^2 - 3x)^(1/3).
(x^2 - 3x)^(1/3)[x + 2(x^2 - 3x)^(4/9)]
Stuck here....

Found 2 solutions by ikleyn, MathTherapy:
Answer by ikleyn(52803)   (Show Source): You can put this solution on YOUR website!
.

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

                A correct factoring is below.


(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.



Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Find the real solutions by factoring.

x(x^2 - 3x)^(1/3) + 2(x^2 - 3x)^(4/3) = 0

I factored out (x^2 - 3x)^(1/3).

(x^2 - 3x)^(1/3)[x + 2(x^2 - 3x)^(4/9)]

Stuck here....

  
 ---- Factoring out GCF, 

Applying the zero-product rule, we get:
                                                                            
                 ---- Cubing each side        OR  
                                                                                                                                                
                     
                              OR      

REAL SOLUTIONS: 

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