I have the answer to this question but I can't understand a part of it. This is the first thing in the exercise:
Let $K$ be a field in $\mathbb{C}$ and $p(x)=x^3+px+q$ a irreducible polynomial in $K[x]$. Consider $x$ a root of $p(x)$ and $D=d^2=-4p^3-27q^2$ the discriminant of $p(x)$. Take an $u=a+bx+cx^2$ of $K(x)$ that it is not in $K$, with minimal polynomial $x^3+a'x^2+b'x+c'$ in $K$. Determine $a'$ and $b'$ in terms of $a, b, c, p, q$.
This part is OK, I did it and undertand it, than there's this:
Show that $K(x)$ is a radical extension of $K$ if and only if $-3D$ is a square in $K$.
The answer I have is this:
So that $K(x)$ is a radical extension of $K$, it is necessary and sufficient that exists an element $u=a+bx+cx^2\in K(x)$ such that $K(x)=K(u)$, $u^3\in K$.
And that is the part I don't get, how do I prove that if $K(x)$ is a radical extension then exists an element $u=a+bx+cx^2\in K(x)$ such that $K(x)=K(u)$, $u^3\in K$, the more confusing part for me is the $u^3\in K$.