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?
 A: It suffices to show any rational root $w$ is an integer $\,n,\,$ by $\,1 = -n(n^2\!+\!1)\,\Rightarrow\, n\mid 1\,$ so $\,n = \pm1,\,$ contradiction.  Suppose $\,w  = c/d\in\Bbb Q.\,$ Note  $d^2$ is a common denominator for all elements $r$ in the ring $\,R = \Bbb Z[w] = \{ a_o + a_1 w + a_2 w^2\ :\ a_i\in\Bbb Z\}.\,$ Thus $\,R\subseteq \Bbb Z/d^2,\,$ i.e. $\,r \in R\,\Rightarrow\,r = n/d^2\,$ for $\,n\in\Bbb Z.\,$ If $\,\color{#c00}{r\not\in\Bbb Z}\,$ then wlog we may assume $\,0 < r < 1\,$ by taking its fractional part - which lies in $\,R\,$ and is nonintegral iff $r$ is. Then 
 $\,r\in \{ 1/d^2,\, 2/d^2,\ldots,(d^2\!-\!1)/d^2\}.\,$  If $r$ is the smallest
element of $R$ in this set then  $r^2$  is an even smaller such element,
since  $\,1 > r > r^2 > 0,\,$ contra minimality of $\,r.\,$  Therefore $\,\color{#c00}{r\in\Bbb Z}.$
A: Gotta hand it to Bill Dubuque.  
Having said this:
Before I read Bill's enlightening answer, and in attempting to steer clear of RRT territory as much as possible, I argued as follows:
As we have seen, with
$x^3 + x + 1 = 0, \tag 1$
and 
$x = \dfrac{p}{q}, \; p, q \in \Bbb Z, \; \gcd(p,q) = 1, \tag 2$
we have
$\dfrac{p^3}{q^3} + \dfrac{p}{q}+ 1 = 0; \tag 3$
which as has been seen leads directly to (upon multiplication by $q^3$)
$p^3 + pq^2 + q^3 = 0; \tag 4$
since $\gcd(p, q) = 1$ we may find $a, b \in \Bbb Z$ such that
$ap + bq = 1; \tag 5$
we multiply by $p^2$:
$ap^3 + bqp^2 = p^2; \tag 6$
from (4),
$q \mid p^3; \tag 7$
then from (6),
$q \mid p^2; \tag 8$
again from (5), this time multiplying by $p$,
$ap^2 + bpq = p; \tag 9$
thus 
$q \mid p; \tag{10}$
so again by virtue of $\gcd(p, q) = 1$:
$q = \pm 1, \tag{11}$
then 
$p^3 + p \pm 1 = 0, \tag{12}$
whence
$p(p^2 + 1) = \pm 1; \tag{13}$
well, 
$p = -1, 0, 1 \tag{14}$
don't solve (13), and if
$\vert p \vert \ge 2, \tag{15}$
then
$p^2 + 1 \ge 5,\tag{16}$
which rules out $p$ as in (15).  Thus (13) has no integer roots, and hence (1) has no rational roots.
A: 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.
