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

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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(52802)   (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:

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....