<s>[INST] Prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational without RRT I need to prove that the roots $\in \Bbb R$ of $x^3+x+1=0$ aren't rational. Obviously, it's easy to use the rational root theorem to prove that there are not rational solutions to this equation, but i want a different approach.
I saw a similar question here, but it didn't provide my "solution".
My try
$x^3+x+1=0$
$x^3+x=-1$
$x(x^2+1)=-1$
Here we have two options (product of two numbers to obtain a negative):
$1.$ $x \gt 0$ $∧$ $x^2+1 \lt 0$ (not possible in $\Bbb R)$
$2.$ $x \lt 0$ $∧$ $x^2+1 \gt 0$ 
So, we use $2.$ to prove that $x \ne 0$, then:
$x(x^2+1)=-1$
$x^2+1=\frac{-1}{x}$
$x^2=\frac{-1}{x}-1$
$x^2=-(\frac{x+1}{x})$
$x=\pm \sqrt {-(\frac{x+1}{x})}$
Then, $\frac{x+1}{x} \ge 0$ because $x \lt 0$ (the equality occurs when $x=-1$, but this doesn't satisfy the original polynomial equation) but here i'm missing the cases $-1 \lt x \lt 0$ (i don't know how to use this to prove that the root is irrational)
This implies that $2$ of the roots are complex, but there are $3$ roots to a third degree polynomial equation.
And here i'm stuck, because i don't know how to prove that the last solution is irrational.
Any hints?
Is there anyway to prove that the last root is irrational?
Is my proof good so far?
 [/INST] If $x$ is real then
$-1 = x(x^2+1)
$
so
$x = \frac{-1}{x^2+1}
$
so
$-1 < x < 0$.
If $x = -c/d$ 
with $(c, d) = 1$,
then
$\frac{c}{d}
= \frac{1}{(c/d)^2+1}
=\frac{d^2}{c^2+d^2}
$
or
$c(c^2+d^2) = d^3$.
If a prime $p$
divides $c$,
then $p | d^3$
so $p | d$,
which contradicts
$(c, d) = 1$.
Therefore
$c = 1$,
so
$1+d^2 = d^3$
or
$1
=d^3-d^2
=d^2(d-1)
$
which can not hold
since it is false for $d = 1$
and
$d^2(d-1) > 1$
for
$d \ge 2$.
Therefore,
there is no rational root.
</s>