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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.

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    $\begingroup$ Here's an equation that fits your requirements: $x^3-3x+1 = 0$ $\endgroup$ Mar 28, 2016 at 19:07
  • $\begingroup$ A cubic equation with 3 distinct solutions, a, b, c, is of the form $d(x- a)(x- b)(x- c)= dx^3- d(a+ b+ c)x^2+d(ab+ ac+ bc)x- abc$ $\endgroup$
    – user247327
    Mar 28, 2016 at 19:09
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    $\begingroup$ You may want to read en.m.wikipedia.org/wiki/Casus_irreducibilis $\endgroup$
    – Macavity
    Mar 28, 2016 at 19:16

4 Answers 4

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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.

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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.

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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 $2\cos\left(\dfrac\pi9\right), 2\cos\left(\dfrac{7\pi}9\right), 2\cos\left(\dfrac{13\pi}9\right)$.

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  • $\begingroup$ I am afraid that this is the only equation that can be found by this method when the leading coefficient is $1$ :( $\endgroup$
    – user65203
    Mar 28, 2016 at 19:52
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Try this

$x^3 - 8x^2 + x +9 = 0$

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