Cubic polynomial with three (distinct) irrational roots I am looking for an equation

$$x^3+ax^2+bx+c=0, \qquad a, b, c \in \Bbb Z,$$

of degree $3$ that has $3$ different roots.
For an equation of degree $2$ it is easy---for example $x^2-2=0$---but I can't find an example for degree $3$.
if there is a method for finding this kind of Equation I will be grateful. 
 A: 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.
A: 
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.
A: Choose $t$ such that $$\cos(3t)=4\cos^3(t)-3\cos(t)=q$$ is a rational number, but not $\cos(t)$.
For instance, with $\cos\left(3\dfrac\pi9\right)=\dfrac12$,
$$4\cos^3(t)-3\cos(t)=\frac12$$ or, after the change of variable $x=2\cos(t)$,
$$\color{green}{x^3-3x-1=0}.$$
The solutions are $2\cos\left(\dfrac\pi9\right), 2\cos\left(\dfrac{7\pi}9\right), 2\cos\left(\dfrac{13\pi}9\right)$.
A: Try this 
$x^3 - 8x^2 + x +9 = 0$
