+0  
 
0
101
2
avatar

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?

 May 30, 2019
 #1
avatar
0

plz

some 1?

 May 30, 2019
 #2
avatar+106519 
+2

       1

________________

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

 

We realize that we can form a difference of cubes if we multiply  the numerator and denominator by

[ 3^(2/3) + 6^(1/3) + 2^(2/3) ]    ....so we have

 

[ 3^(2/3) + 6^(1/3) + 2^(2/3) ]

______________________________________  =

[ 3^(1/3) - 2^(1/3) ] [ 3^(2/3) + 6^(1/3) + 2^(2/3)]

 

(3^2)^(1/3) + 6^(1/3) + (2^2)^3

_________________________  =

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

 

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

_____________________  =

          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

 May 30, 2019

11 Online Users

avatar