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.
 A: Let $a,b,c$ be the three real roots of $p(x) = x^3-3x+1$.
Consider the interpolating polynomials $f,g$ such that $f(a)=b, f(b)=c, f(c)=a$ and $g(a)=c, g(b)=a, g(c)=b$.
With some Galois theory, one can show that they are defined over $\Bbb Q(\sqrt \Delta)$ (and since all roots are real, the polynomials are real, so $\Delta > 0$). In fact since here $\Delta$ is a square, they have rational coefficients.
It turns out that $f,g$ are the polynomials $x \mapsto x^2-2$ and $x^2 \mapsto -x^2-x+2$. It's a bit hard to find, but easy to check because a straightforward computations shows that if $a$ is a root, then so are $f(a)$ and $g(a)$.
Well this tell us that if $K$ is any subfield of $\Bbb R$ then if $p(x)$ has a root in $K$ then the other two roots are also in $K$. So either $p(x)$ is irreducible or it has three linear factors.

Now, if $K \subset L = K(\sqrt[n] y) \subset \Bbb R$ with $y \in K$ and $n$ prime, then if $p(x)$ splits in $L$ then it was already split in $K$ :
Since $n$ is prime, there is no intermediate subfield between $K$ and $L$, so if $p$ splits in $L$ and is irreducible over $K$ then $L = K(a)$ and $n=3$. 
Then $L$ is Galois over $K$ which is impossible because cube root extensions are not Galois :
If $q(a)$ is a root of $Y^3-y$ for some polynomial $q \in K[X]$, then $Y^3-y$ splits into $(Y-q(a))(Y-q(b))(Y-q(c))$ over $K(a)$ (remember that $b=f(a)$ and $c=g(a)$), and we get $(q(a) -q(b))^2 + 3q(c)^2 = 0$, from which we get $q(c)=0$ then $y=0$, and $K=L$, contradiction.

Now if you had a way to obtain $a$ from rationals with the usual operations and taking real roots, then there would be a finite tower of radical extensions $\Bbb Q \subset K_1 \subset \ldots \subset K_n \subset \Bbb R$ with $a \in K_n$. With the above result, if $a \in K_n$ then $a \in K_{n-1}$ and so on, which implies that $a \in \Bbb Q$ in the first place.
Since the roots are not rationals, it is impossible to obtain them from rationals with those operations.
A: The discriminant $\Delta=(-4\times-27)-27=3\times 27=81.$ Since the discriminant is not a perfect cube, then the polynomial is not factorable because of the formula to the cubic equation.
But I am not quite sure about this. The discriminant is given by the formula $\Delta=18abcd-4b^3d+b^2c^2-4ac^3-27a^2d^2$.
Here's about the discriminant.
A: A good place to start for rational numbers is Eisenstein's criterion. This following link will tell you more about it:
https://en.wikipedia.org/wiki/Eisenstein%27s_criterion
A: See "What Is Mathematics?" by Courant and Robbins, ch. 3. If you take $x=2\cos t$ then you'll get an equation $\cos 3t=-1/2$. It correspond to trisection of the angle $120^\circ$, which is impossible (see Courant - Robbins again).
A: Let us define our cubic as:
$$f(x)=ax^3+bx^2+cx+d=0$$
We consider the polynomial:
$$D(x)=3ax^2+2bx+c$$
We test for when $D(x)=0$, say $x_1,x_2$.
If $x_1,x_2$ don't exist, then $f$ has only one real root.
If $x_1,x_2$ both exist, and $f(x_1),f(x_2)$ have different signs, then $f$ has three real roots.
If either or both of $f(x_1),f(x_2)$ are zero, then $f$ has three real roots, two (or all three if both) being the same.
Otherwise, i.e. $\operatorname{sign}(f(x_1))=\operatorname{sign}(f(x_2))$, there is only one real root.
A: 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 $\mathbb{R-Q}$.
