Rationalize the denominator of \(\displaystyle \frac{1}{\sqrt[3]{3} - 1}\). 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 ?
+128408 Multiply top / bottom by 1 + 3^(1/3) + 3^(2/3)
Note that the bottom becomes
(3^(1/3) - 1) ( 1 + 3^(1/3) + 3^(2/3) ) =
3^(1/3) + 3^(2/3) + 3 - 1 - 3^(1/3) - 3^(2/3) = 2
So we have
1 + 3^(1/3) + 3^(2/3) ∛1 + ∛3 + ∛9
__________________ = _____________
2 2
A + B + C + D = 15
