SOLUTION: I am having some difficulties working functions which have numbers raised to fractional powers, like the following question. Solve: x^(2/3) - 2x^(1/3) - 15 = 0 I understand u

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Question 277573: I am having some difficulties working functions which have numbers raised to fractional powers, like the following question.
Solve: x^(2/3) - 2x^(1/3) - 15 = 0
I understand using a placeholder thing, reorganizing the problem, and then solving for the placeholder, but I am having problems with raising to the fractional powers while solving for X. Could you explain how please?
I got this question solved down to +u%5E2+-+2u+-15+=+0+ where u = x^(1/3) The possible solutions for u are 5, and -3. I need help solving for X now.
Thanks
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since u=x%5E%281%2F3%29%5E%22%22, we can convert to radical notation to get u=root%283%2Cx%29. It's just a different way to same thing.


Now because the possible solutions for 'u' (no pun intended) are 5 and -3, this means that u=5 or u=-3


Let's find the solution(s) in terms of 'x' when u=5


u=root%283%2Cx%29 Start with the given equation.


5=root%283%2Cx%29 Plug in u=5


%285%29%5E3=%28root%283%2Cx%29%29%5E3 Cube both sides to undo the cube root.


%285%29%5E3=x Cube the cube root of 'x' to just get 'x'.


125=x Cube 5 to get 125.


So one solution in terms of 'x' is x=125

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Now let's find the solution(s) in terms of 'x' when u=-3


u=root%283%2Cx%29 Start with the given equation.


-3=root%283%2Cx%29 Plug in u=-3


%28-3%29%5E3=%28root%283%2Cx%29%29%5E3 Cube both sides to undo the cube root.


%28-3%29%5E3=x Cube the cube root of 'x' to just get 'x'.


-27=x Cube -3 to get -27.


So another solution in terms of 'x' is x=-27


So the solutions in terms of 'x' are x=125 or x=-27