+1347 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 $A + B + C + D$?
+128731 \( $\displaystyle \frac{1}{2 - \sqrt[3]{2}}$ \)
\($\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$ \)
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
