1. Powered fractions

What are the steps to solve this problem?

x^(2/3) - x^(1/3) - 6 = 0

2. This is actually a quadratic equation in $x^{1/3}$.

Recalling the laws of indices we can say that $x^{2/3} = \left(x^{1/3}\right)^2$

To illustrate the point let $u = x^{1/3}$ - thus we have $u^2-u-6=0$

Solve the quadratic using your favourite method (this one does factorise) to get two values of u.

Once you have the values of u you put $x^{1/3} = root_1$ and $x^{1/3} = root_2$ where root_1 and root_2 are your values of u. It should then be straightforward to find x

3. Originally Posted by e^(i*pi)
Once you have the values of u you put $x^{1/3} = root_1$ and $x^{1/3} = root_2$ where root_1 and root_2 are your values of u. It should then be straightforward to find x

I don't understand that part. I got the two values of u which were -2 and 3.

4. Originally Posted by Joaco
I don't understand that part. I got the two values of u which were 2 and 3.
In that case you have u= 2 and u=3

Since we defined u as being equal to x^{1/3} then x^{1/3} must equal 2 and 3.

$u = x^{1/3} = 2$ and $u = x^{1/3} = 3.$

You can then solve for x, you will get two solutions but this simply means two values of x satisfy this equation

5. I'm sorry but how do you solve for x?

EDIT: Nevermind I wasn't thinking :P

So I would get x= 8 or x=-27

Thank you for your help !