+84 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 \(A + B + C + D\)?
+934 You'll need to combind conjugates with rationalizing a cube root.
+23252 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
