Precalculus: Find the inverse of 1 + (2)^(1/3)

I did the conjugate and got this:

1/(1+2^(1/3)) = (1 - 2^(1/3))/(1 - 2^(2/3))

but its not simplified...does this have anything to do with (2)^(1/3) being a root of p(x) = x^3 - 2 ???

Use the sum of cubes formula
a^3 + b^3 = (a + b)(a^2 - ab + b^2) with a = 1 and b = 2^(1/3):

So, 1/(1 + 2^(1/3))
= 1 * (1 + 2^(1/3) + (2^(1/3))^2) / [(1 + 2^(1/3)) * (1 + 2^(1/3) + (2^(1/3))^2)]
= (1 + 2^(1/3) + 4^(1/3)) / (1 + 2), via sum of cubes
= (1/3)(1 + 2^(1/3) + 4^(1/3)).

I hope this helps!