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?

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.
