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
