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?

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.
