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Rationalize the denominator of \(\displaystyle \frac{1}{\sqrt[3]{3} - \sqrt[3]{2}}\). After formatting your answer in the form \(\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}\), and simplifying the fraction down to its lowest terms, what is \(A + B + C + D\)?

 Apr 8, 2020
 #1
avatar+930 
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See if this helps.

 Apr 8, 2020
 #2
avatar+930 
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You'll need to combind conjugates with rationalizing a cube root.

HELPMEEEEEEEEEEEEE  Apr 8, 2020
 #3
avatar+20967 
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You can use this identity:  a3 - b3  =  (a - b)(a2 + a·b + b2)

 

In this case:  a = cubr(3)  and  b = cubr(2)

 

So:  multiply the numerator and denominator by:  cubr(32) + cubr(3·2) + cubr(22}  =  cubr(9) + cubr(6) + cubr(4)

 

The numerator becomes:  cubr(9) + cubr(6) + cubr(4)

while the denominator becomes 3 - 2  =  1

 

A + B + C + D  =  cubr(9) + cubr(6) + cubr(4) + 1

 Apr 8, 2020

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