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Rationalize the denominator of $\frac{3}{2\sqrt[3]{5}}$. The answer can be written in the form of $\frac{A\sqrt[3]{B}}{C}$, where $A$, $B$, and $C$ are integers, $C$ is positive, and $B$ is not divisible by the cube of any prime. Find $A+B+C$.

 Jun 15, 2019
 #1
avatar+104962 
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\($\frac{3}{2\sqrt[3]{5}}$ \)   =     

 

        3                  [ 5^(2/3) ]

_________         ________   =

2 * (5)^(1/3)         [5^(2/3) ]

 

 

3 [ 5^(2/3) ]                  3 *(5^2)^(1/3)              3 ∛25

___________   =        ____________  =       ______

    2 * 5                               10                            10

 

 

A = 3   B  = 25    C = 10

 

And their sum  =   38

 

 

cool cool cool

 Jun 15, 2019

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