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Rationalize the denominator of \(\displaystyle \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}}\)  . 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?

tertre  May 29, 2017
 #1
avatar+90001 
+2

We can write

 

1 / [ 3^(1/3) - 2 ^(1/3) ]

 

Multiply top/bottom  by   [ 3^(2/3) +  (3*2)^(1/3) + 2^(2/3) ]

 

And we have

 

[ 3^(2/3) + (3*2)^(1/3) + 2^(2/3) ]  /  [ 3 + (18)^(1/3) + (12)^(1/3) - (18)^(1/3) - 12^(1/3) - 2 ] 

 

[ (9)^(1/3)  + (6)^(1/3) + (4)^(1/3) ]  /  1

 

So   A + B + C + D    =  9 + 6 + 4 + 1  =   20

 

 

 

cool cool cool

CPhill  May 29, 2017

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