Since a3 + b3 = (a + b)(a2 - ab + b2)
Use (a2 - ab + b2) where a = cuberoot(3) and b = cuberoot(2).
\(\dfrac{1}{\sqrt[3]{3}+\sqrt[3]{2}}\\ =\dfrac{(\sqrt[3]{3})^2-\sqrt[3]{3\times 2}+(\sqrt[3]{2})^2}{3 + 2}\\ =\dfrac{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}{5}\)
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