<s>[INST] Proving that I cannot factor a cubic polynomial with Algebra alone How do I prove a given polynomial is not factor-able using only radicals, basic arithmetic operations, and real, rational numbers?  For example, how can I prove that
$$0=x^3-3x+1$$
Has no solution with the given restraints?


Persistently, I want an algebraic (pre-calc. type of algebra) proof.

An idea I've had is to assume it does have one solution expressible with the given and then disproving using contradictions, but it seems be fruitless.
 [/INST] So we have that $$x^3-3x+1=0 \tag1$$
We can rewrite this as:
$$\begin{align} & x^3+3x^2+3x+1=3x^2+6x \\
\implies & (x+1)^3=3x(x+2)=3\{(x+1)-1\}\{(x+1)+1\}=3(x+1)^2+2(x+1)-1 \\
\implies & (x+1)^3-3(x+1)^2-2(x+1)+1=0 \\
\implies & (x+1)^3-3(x+1)^2+3(x+1)-1=5(x+1)-2 \\
\implies & (x+1-1)^3=5(x+1)-2 \\
\implies & x^3=5x+3 \\
\implies & x^3-5x-3=0 \tag2\end{align} $$
Hence, we see that we can re-arrange $(1)$ to get $(2)$.
Now, $(1)-(2) \implies 5x+3-3x+1=0 \implies x=-2$
So $x=-2$ is a common solution of both $(1)$ and $(2)$. But on verification, we see that $x=-2$ does not satisfy either equation.
This is a clear implication that $(1)$ has no solution in $\mathbb{Q}$. 
On application of Cardan's Theorem, we get that the equation, being a cubic one has all $3$ solutions in $\mathbb{R-Q}$.
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