SOLUTION: I am having so much trouble with this question. I will show you the original question and then will show what I have. Please complete the problem for me from start to finish so t

Question 134716: I am having so much trouble with this question. I will show you the original question and then will show what I have. Please complete the problem for me from start to finish so that I am no longer as confused as I am now. This is what I have:
(3x)^-1/3=
3^-1/3 * x^-1/3=
1/3^1/3 * 1/x^-1/3=
3^1/3= 3cubert3=
1/3cubert3x
I am not allowed to leave radicals in the denominator, I am completely confused, can you please solve the original? The original problem again is (3x)^-1/3

Found 2 solutions by stanbon, vleith:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
(3x)^(-1/3)
= 1/(3x)^(1/3)
Multiply numerator and denominator by (3x)^(2/3) to get:
= [(3x)^(2/3)]/3x

======================
Cheers,
stan H.

Answer by vleith(2983) (Show Source): You can put this solution on YOUR website!
Given:

Next, clear the denominator of any radicals
Multiply by which is 1

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