$4qx^3-3qx-p$ is not reducible over $\mathbb{Q}$ Show $4qx^3-3qx-p$ is irreducible over $\mathbb{Q}$, where $\gcd(p,q)=1$, $q>1$, and $q$ is cubefree.
This question is closely related to trisecting an angle is impossible under certain conditions.
 A: If $f(x) := 4qx^3 - 3qx - p$ is reducible over $\mathbb{Q}$, then it has a rational root $a/b$ (and WLOG $b \in \mathbb{N}$, $\text{gcd}(a,b) = 1$). By the rational root theorem, $a \mid p$, $b \mid 4q$. We have $4q(\frac{a}{b})^3 - 3q(\frac{a}{b}) - p = 0$, and to bring the middle term to $\mathbb{Z}$, we multiply by $4$, to get $16q(\frac{a}{b})^3 = 12q(\frac{a}{b}) + 4p$. Then the RHS is an integer, so $b^3 \mid 16qa^3 \implies b^3 \mid 16q \implies b \mid 2$. Since $f(x)$ has no integer roots (since $\frac{p}{q} \not \in \mathbb{Z}$), $b = 2$, but then $4q(\frac{a}{2})^3 - 3q(\frac{a}{2}) - p = 0$, so $q(a^3 - 3a) = 2p$. But $a^3 - 3a$ is always even, so $p = q(\frac{a^3 - 3a}{2})$, contradicting $\text{gcd}(p,q) = 1$.
As you noted, this shows exactly that if $\cos(\theta) = \frac{p}{q}$ ($p$ and $q$ as above), then $\theta$ is not trisectable. 
A: If it's reducible, it must have a root, which root we'll write as $r/s$, a quotient of two integers in lowest terms. So $$4qr^3-3qrs^2-ps^3=0$$ Now let $t$ be a prime dividing $q$. Then $t$ must divide $ps^3$, but $\gcd(p,q)=1$, so $t$ divides $s^3$, so $t$ divides $s$, so $t^3$ divides $s^3$. Then $t^3$ also divides $3qrs^2$, so it divides $q$. This contradicts the assumption on $q$, so $q$ is not divisible by any prime; we must have $q=\pm1$. 
Can you take it from there?
