# 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.