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.

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 $\mathbb{R-Q}$.
