Start by assuming that $\rm\:a/b\:$ is a root in lowest terms, then follow the hints. Since $\rm\:a/b\:$ is in lowest terms, $\rm\:a,b\:$ are not both even. Next, considering the remaining parity cases for $\rm\:a,b\:$ leads to the contradiction that $\rm\: a^3 + a\:b^2 + b^3 = 0\:$ is odd, since if $\rm\:a,b\:$ have opposite parity then the sum has two even terms and one odd term, and if $\rm\:a,b\:$ have equal parity then they are both odd, so all three summands are odd, hence so too is their sum. This parity contradiction completes the proof.

If you know a little algebra, below is a generalization from my post here. (e.g. let $\rm\:R = \mathbb Z\:$ below).

**PARITY ROOT TEST** $\ $ Suppose $\rm\:f(x)\:$ is a polynomial with coefficients in a ring $\rm R$ with parity. Then $\rm\:f(x)\:$ has no roots in $\rm R$ if $\rm\:f(x)\:$ has constant coefficient and coefficient sum both being odd. Further, $\rm\:f(x)\:$ has no roots in the fraction field of $\rm R$ when $\rm R$ is a domain such that $\rm\:R/\hat{2}\:\!\cong\mathbb Z/2,\:$ and $\rm\:f\:$ has odd lead coefficient, and $0$ is the only $\rm\:r\in R\:$ divisible by unbounded powers of $\hat 2$.

**Proof** $\ $ The test simply verifies $\rm\ f(0) \equiv f(1) \equiv 1\:\ (mod\ 2),\ $ i.e. $\rm\: f(x)\:$ has no roots $\rm\: (mod\ 2)\:.\ $ Therefore $\rm\:f(x)\:$ has no roots in $\rm\:R\:.\:$ For the fractional case, by the hypothesis, we may cancel powers of $\hat 2$ from a fraction $\rm\:a/b\:$ till $\rm\:a,b\:$ aren't both even. $\rm\:b\:$ isn't odd, else $\rm\:a/b\equiv a\pmod{2}\:$ would be a root, so $\rm\:b\:$ is even and $\rm\:a\:$ is odd. Now $\rm\:0 = b^n\:f(a/b) = f_n a^n + b\:(\cdots) \equiv 1\pmod{2},\:$ contradiction, since the leading coefficient $\rm\:f_n$ and $\rm\:a\:$ are both odd, and $\rm\:b\:$ is even. $\ $ **QED**