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

Algebra ->  Radicals -> 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      Log On


   

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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
Given: %283x%29%5E%28-1%2F3%29
1%2F%283x%29%5E%281%2F3%29
Next, clear the denominator of any radicals
Multiply by %283x%29%5E%282%2F3%29%2F+%283x%29%5E%282%2F3%29 which is 1
%283x%29%5E%282%2F3%29%2F%28%283x%29%5E%281%2F3%29%2A%283x%29%5E%282%2F3%29%29
%283x%29%5E%282%2F3%29%2F%283x%29%5E%28%281%2F3%29+%2B+%282%2F3%29%29
%283x%29%5E%282%2F3%29%2F%283x%29