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)

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

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 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) (Show Source): You can put this solution on YOUR website!
Since , we can convert to radical notation to get . 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 or

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

Start with the given equation.

Plug in

Cube both sides to undo the cube root.

Cube the cube root of 'x' to just get 'x'.

Cube 5 to get 125.

So one solution in terms of 'x' is

--------------------

Now let's find the solution(s) in terms of 'x' when

Start with the given equation.

Plug in

Cube both sides to undo the cube root.

Cube the cube root of 'x' to just get 'x'.

Cube -3 to get -27.

So another solution in terms of 'x' is

So the solutions in terms of 'x' are or
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