help how to rationalize denominator

To rationalize an expression such as $\frac{1}{2 - \sqrt[3]{2}}$, you first recognize that there is a cube root in the denominator. This immediately provides insight as to how to approach this particular problem. Personally, I find working with fractional exponents is far easier than radicals, so I will almost always convert to this.

Since there is a cube root within the denominator, we will have to make the denominator a perfect cube in order to ensure that the denominator has no radicals. We can do this by making a key observation: $A^3 - B^3 = (A - B)(A^2 + AB + B^2)$. This is the formula for the difference of a perfect cube. This information should be enough to rationalize this problem and ones like it. 

If we look closely at the denominator, we can match up the formula for a perfect cube to the denominator of this particular problem.

If we let $A = 2$ and $B = \sqrt[3]{2} = 2^\frac{1}{3}$, then that would make the following identity. 

$A = 2; B = 2^\frac{1}{3} \\ 2^3 - \left(2^\frac{1}{3}\right)^3 = \left(2 - 2^\frac{1}{3}\right)\left(2^2 + 2*2^\frac{1}{3} + \left(2^\frac{1}{3}\right)^2\right) \\ 6 = \left(2 - 2^\frac{1}{3} \right) \left(4 + 2^\frac{4}{3} + 2^\frac{2}{3}\right) \\$

Now, we can rationalize because we have the term that will cancel out all radicals within the denominator.

$\frac{1}{2 - 2^\frac{1}{3}} * \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{4 + 2^\frac{4}{3} + 2 ^\frac{2}{3}} = \frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6}$

Now, we have to convert to the radical notation and figure out how it matches the desired format of $\displaystyle \frac{\sqrt[3]{A} + \sqrt[3]{B} + \sqrt[3]{C}}{D}$.

$\frac{4 + 2^\frac{4}{3} + 2^\frac{2}{3}}{6} = \frac{4 + \sqrt[3]{2^4} + \sqrt[3]{2^2}}{6} = \frac{\sqrt[3]{64} + \sqrt[3]{16} + \sqrt[3]{4}}{6}$

Now, we identify the desired variables and find their sum.

$A = 64, B=16, C=4, D=6 \\ A + B + C + D = 64 + 16 + 4 + 6 = 90$