Show that the equation $\frac{1}{2} =4x^3-3x$ has no rational root. Suppose it has rational root, then $x=\frac{p}{q}$, where $q\neq 0$ and $(p, q)=1$.
Then the equation can be written as,
$$q^3=2(4p^3-3pq^2)$$
So $q$ is even, this force $p$ to be an odd.
So we can substitute $p=2k+1$ and look for any contradiction.
Is the way correct? Is there any easy way to do it?
 A: You can apply the Rational root theorem to the equation
$$
8x^3 - 6x -1 = 0 \, .
$$
It states that if $x=p/q$  with integers $p, q$ is a rational solution of that equation then $p$ is a factor of $a_0 = 1$, and $q$ is a factor of $a_3 = 8$.
That gives (a small number of) candidates for rational roots which you can exclude by substituting them into the equation.
A: It is quite easy to solve with the rational root theorem and others have already mentioned it. For similar problems, you may be interested also in Eisenstein's criterion. https://en.wikipedia.org/wiki/Eisenstein%27s_criterion .
Rewriting your equation as $8x^3 -6x -1 = 0$ you'll see that Eisenstein's criterion is not applicable. we may "fix" that by substituting $x$ by $x+1$. Obtaining:
$$ Q = 8X^3 + 24X^2 - 30X -3$$
Here the prime $p = 3$ satisfies the criterion and as such $Q$ is irreducible over $\mathbb{Q}[X]$ because our shift was an automorphism on $\mathbb{Q}[X]$, Q is irreducible iff $8X^3 -6X -1$ is.
A: Write $x=\frac p q$ with $(p,q)= 1$ and $p>0$. Then you can write $$q^3 = 2p(4p^2-3q^2).$$
This shows that $p | q^3 \implies p=1$, giving $$q^2(q+6)=8.$$
Since $q,q^2|8$ we can only have $q=\pm 1, \pm 2$, but none of them satisfy the last equation.
