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?
+128473 1
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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
