Cubic equation $X^3+pX+q$ not solvable by radicals if $D=-4p^3 - 27q^2 >0$

Question

How can one prove that the real cubic equation $$P(X)=X^3+pX+q$$ is not solvable by real radicals when $$D=-4p^3 - 27q^2 >0?$$

Which means that there is no sequence of extension: $$\mathbb R=L_0 \subset L_1 \subset ... \subset L_n=L$$ with $a\in L$ root of $P$ and for $0 \leqslant i \leqslant n-1$, $L_{i+1}= L_i(u_i)$, where $u_i^{p_i} \in L_{i-1}$, $p_i$ being a prime number, and $u_i$ a stricly positive real.

Answer 1

Look at https://en.wikipedia.org/wiki/Cubic_function#The_nature_of_the_roots. Your question makes no sense cubics are solvable by radicals for reference look at the linked wiki page.

But if you mean id $D>0$ then there are three real roots then you can do this Hint:

First observe that if $D >0$ then $p<0$ then the equation $y=x^3+px+q$ has two local maxima. At the points $\sqrt{-p/3}$ and $-\sqrt{-p/3}$. Show that at these two points the values of $y$ have opposite signs and hence the curve must have crossed the $X$ axis three times ( note as $x$ goes to $-\infty$ (or $\infty$) $y$ goes to $-\infty $ (or $\infty$)).

Answer 2

The cubic $$x^3+px+q=0$$ is always soluble by radicals. With the classical Cardano formula we have: $$ x=u+v=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}} $$ with the condition $uv=-p/3$.

Simply, if $27q^2+4p^3 <0$, we have to use imaginary numbers to evaluate the square root, and this was the great discovery of Cardano.

If $27q^2+4p^3 \ge 0$ this solution use only real radicals.