Rational Numbers – Show $ (3 + sqrt {2}) ^ {2/3} $ is irrational

I am asked to prove that $ (3+ sqrt {2}) ^ {2/3} $ is irrational about the set of rational zeros.

I have so far:

$ x = (3+ sqrt {2}) ^ {2/3} $

$ x ^ 3 = (3+ sqrt {2}) ^ {2} $

$ x ^ 3 – 11 – 6 sqrt {2} = 0 $

From here I do not know how to get it into the RZT-approved form. The only way I can proceed from is to say every reasonable solution to the form $ r = frac {c} {d} $ $ c, d, in mathbb {Z} $ would have to have $ d = pm $ 1 and $ c $ divide $ -11 -6 sqrt {2} $,

so we have

$ -11 -6 sqrt {2} = z c $, $ z in mathbb {Z} $

$ -6 sqrt {2} = -11 + zc $

Which clearly no $ c $ will satisfy. So it's proven, but I feel like I did not use the RZT the way I should. At least it's different from the other example problems I've made, for example, to prove $ sqrt {3} $ is irrational.