Cubic polynomial with three (distinct) irrational roots

if there is a method for finding this kind of Equation I will be grateful.

Let $f(x)=x^3+ax^2+bx+c$.

What you are looking for is the set $(a,b,c)\in\mathbb Z$ such that $$-4b^3-27c^2+a^2b^2+18abc-4a^3c\gt 0$$ (see discriminant) and $$f(\pm d)\not=0\quad\text{where $\quad d$ is a factor of $c$}$$ (see rational root theorem)

To find (infinitely many) concrete examples easily, let us set $b=0,c=1$.

We are looking for $a\in\mathbb Z$ such that $$4a^3+27\lt 0,\qquad a\not=-2,\qquad a\not=0.$$

Hence, we can see that $$\color{red}{x^3+ax^2+1=0\qquad \text{where $\quad a\le -3$}}$$ has three distinct irrational roots.

By the Rational Root Theorem, any rational root of $p(x) = x^3 + a x^2 + b x + c$ is given by $\pm d$, where $d$ is a factor of $c$. So, if we set $c = \pm 1$, if $p$ has a rational roots it must be $\pm 1$.

On the other hand, if we choose the signs of the coefficients of $p$ to alternate, then Descartes' Rule of Signs gives that all of the roots are positive. So, if we pick the coefficients of $a$ and $b$ and $c = \pm 1$ to alternate, then $x = 1$ is the only possible rational root, and thus $p$ has no rational roots if $p(1) = 1 + a + b + c \neq 0$.

For example, if we take $a = 0, b < 0, c = 1$, then we deduce that $$p(x) = x^3 + b x + 1$$ has no rational roots provided $b \neq -2$. Now, $p(0) = 1$, so if $p(1) < 0$---that is, if $b < -2$, the asymptotic behavior of $p$ and the Intermediate Value Theorem imply that $p$ has three real, irrational roots (one in each of $(-\infty, 0), (0, 1), (1, \infty)$). Notice that this family, $$\color{#df0000}{\boxed{p(x) = x^3 + b x + 1, \qquad b < - 2}},$$ consists precisely of the reciprocal polynomials of the polynomials in the family in mathlove's good answer.