Rationalizing Denominators

$\dfrac{1}{\sqrt[3]{3}+\sqrt[3]{2}}\\ =\dfrac{(\sqrt[3]{3})^2-\sqrt[3]{3\times 2}+(\sqrt[3]{2})^2}{3 + 2}\\ =\dfrac{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}{5}$

Since  a³ + b³  =  (a + b)(a² - ab + b²)

Use  (a² - ab + b²)  where  a = cuberoot(3)  and  b = cuberoot(2). 

1 / [3^(1/3) + 2^(1/3) ] 

We realize that the denominator can be made to be a sum of cubes if we multiply  numerator and denominator by  [3^(2/3) - 3^(1/3)*2^(1/3) + 2^(2/3) ]

[3^(2/3) - 3^(1/3)*2^(1/3) + 2^(2/3) ] / ( [3^(1/3) + 2^(1/3) ]  [3^(2/3) - 3^(1/3)*2^(1/3) + 2^(2/3) ] )

[3^(2/3) - 3^(1/3)*2^(1/3) + 2^(2/3) ] /  [ (3^(1/3))^3  + (2^(1/3))^3 ]  =

[ 3^(2/3) - 3^(1/3)*2^(1/3) + 2^(2/3) ] / 5  =

[3^(2/3) - 6^(1/3) + 2^(2/3) ] / 5  =

[ ∛9 - ∛6  +  ∛4 ]  / 5