<s>[INST] prove that $f(x) = x^3 + 3x - 1$ is irreducible in $\mathbb Q[X]$ 
Prove that $f(x) = x^3  + 3x - 1$ is irreducible in $\mathbb Q[X]$.
Let $\theta$ be a root of $f(x)$. Compute $\frac{1}{\theta}$ and $(2 + \theta^2)^{-1} $ in $\mathbb Q[\theta ]$.

\begin{array}{l}
 f\left( \theta  \right) = \theta ^3  + 3\theta  - 1 = 0 \\ 
  \Leftrightarrow \theta \left( {3 + \theta ^2 } \right) = 1 \\ 
  \Leftrightarrow \frac{1}{\theta } = \left( {3 + \theta ^2 } \right);\left( {\theta  \ne 0} \right) \\ 
 \end{array}
\begin{array}{l}
 \frac{1}{\theta } = 3 + \theta ^2 ;\left( {\theta  \ne 0} \right) \\ 
  \Leftrightarrow \frac{1}{\theta } - 1 = 2 + \theta ^2  \\ 
  \Leftrightarrow \left( {\frac{1}{\theta } - 1} \right)^{ - 1}  = \left( {2 + \theta ^2 } \right)^{ - 1} \quad ;\left( { \pm \sqrt 2  \notin Q} \right) \\ 
  \Leftrightarrow \left( {2 + \theta ^2 } \right)^{ - 1}  = \frac{\theta }{{1 - \theta }}\quad ;\left( {\theta  \ne 1} \right) \\ 
 \end{array}
But I can't show that $f$ is irreducible.
 [/INST] Another idea is that if $f(x)$ is factorable over $\Bbb Q[x]$ it must have at least one rational zero. However, by the rational zero theorem, $\pm 1$ are the only possible rational zeros, but neither one is a zero.
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