# 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 ->  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

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 Algebra: Radicals -- complicated equations involving roots Solvers Lessons Answers archive Quiz In Depth

 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(57352)   (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(2825)   (Show Source): You can put this solution on YOUR website!Given: Next, clear the denominator of any radicals Multiply by which is 1