I've seen a lot of questions like mine around here but most of them use the rational root theorem. However, I am not allowed to use this theorem since I haven't seen it in class. In fact, I should be able to solve this problem only by analyzing parities. Here's what I have so far:
Suppose, by contradiction, that there is at least one rational root and it is of the form $\frac{p}{q}$, $p$ and $q$ are coprimes and $q\neq0$. So we have $(\frac{p}{q})^{2018}-2(\frac{p}{q})^3+24=0$ and we can divide both sides of the equation by $q^{2018}$. Then we are left with $p^{2018}-2p^3q^{2015}+24q^{2018}=0$. Note that $2p^3q^{2015}$ and $24q^{2018}$ are always even
Case 1: $p$ and $q$ odd
This cannot be the case since odd - even + even cannot yield an even result.
Case 2: $p$ and $q$ even
$p$ and $q$ are coprime, thus this also cannot happen.
Case 3: $p$ odd and $q$ even
Similar to the 1st case, also cannot happen.
Case 4: $p$ even and $q$ odd
This is the only possible case, since even - even + even can result zero.
I should get a contradiction, but I cannot see how. Any tips? Is this the right way to prove this?
Hint
$p^{2018}=2(p^3q^{2015}-12q^{2018})$, so $p$ is even and then we can write $p=2k$.
$2^{2018}k^{2018}=2^4k^3q^{2015}-24q^{2018}\to 2^{2015}k^{2018}-2k^3q^{2015}=3q^{2018}$
Looking the last equation, conclude that $q$ is also even. Which is a contradiction.