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Rationalize the denominator of $\displaystyle \frac{1}{2 - \sqrt[3]{2}}$. With your answer in the form $\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$, and the fraction in lowest terms, what is $A + B + C + D$?

 Sep 26, 2023
 #1
avatar+128826 
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\( $\displaystyle \frac{1}{2 - \sqrt[3]{2}}$ \)

 

\($\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$ \)

 

Note that

 

(2^3 - 2) =  (2 - 2^(1/3)) ( 4 + 2^(4/3) + 2^(2/3) )

 

So....multiply  numerator / denominator   by   ( 4 + 2^(4/3) + 2^(2/3) )  and we  have

 

4 + 2^(4/3) + 2^(2/3)             ∛64  + ∛16 + ∛4                

________________    =    ____________________    

     2^3  - 2                                          6

 

A  + B + C + D =   64  + 16 + 4  + 6  =      90 

 

 

cool cool cool

 Sep 27, 2023

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