proof of casus irreducibilis for real fields
The classical statement of the casus irreducibilis^{} is that if $f(x)$ is an irreducible^{} cubic polynomial with rational coefficients^{} and three real roots, then the roots of $f(x)$ are not expressible using real radicals^{}. One example of such a polynomial^{} is ${x}^{3}3x+1$, whose roots are $2\mathrm{cos}(2\pi /9),2\mathrm{cos}(8\pi /9),2\mathrm{cos}(14\pi /9)$.
This article generalizes the classical case to include all polynomials whose degree is not a power of $2$, and also generalizes the base field^{} to be any real extension^{} of $\mathbb{Q}$:
Theorem 1.
Let $F\mathrm{\subset}\mathrm{R}$ be a field, and assume $f\mathit{}\mathrm{(}x\mathrm{)}\mathrm{\in}F\mathit{}\mathrm{[}x\mathrm{]}$ is an irreducible polynomial^{} whose splitting field^{} $L$ is real with $F\mathrm{\subset}L\mathrm{\subset}\mathrm{R}$. Then the following are equivalent^{}:

1.
Some root of $f(x)$ is expressible by real radicals over $F$;

2.
All roots of $f(x)$ are expressible by real radicals over $F$ using only square roots;

3.
$F\subset L$ is a radical extension;

4.
$[L:F]$ is a power of $2$.
Proof. That $(2)\Rightarrow (1)$ is obvious, and $(3)\Rightarrow (1)$ since $F\subset L$ is radical, and is real since $L\subset \mathbb{R}$. $(4)$ implies that $G=\mathrm{Gal}(L/F)$ has order a power of $2$. Since $G$ is a $2$group, it has a nontrivial center (this follows directly from the class equation^{}, or look here (http://planetmath.org/ANontrivialNormalSubgroupOfAFinitePGroupGAndTheCenterOfGHaveNontrivialIntersection)) and thus has a normal subgroup^{} $H$ of order $2$, which corresponds to a subfield^{} $M$ of $L$ Galois over $F$ with $[L:M]=2$. But then $\mathrm{Gal}(M/F)$ is also a $2$group, so inductively we see that we can write
$$F={K}_{0}\subset {K}_{1}\subset \mathrm{\dots}\subset {K}_{m1}=M\subset {K}_{m}=L$$ 
where $[{K}_{i}:{K}_{i1}]=2$. Thus each ${K}_{i}$ is obtained from ${K}_{i1}$ by adjoining a square root; it must be a real square root since $L\subset \mathbb{R}$. This shows that $(4)\Rightarrow (2)$ and $(3)$.
The meat of the proof is in showing that $(1)\Rightarrow (4)$. Let the roots of $f(x)$ be ${\alpha}_{1},\mathrm{\dots},{\alpha}_{m}$, and assume, by renumbering if necessary, that $\alpha ={\alpha}_{1}$ lies in a real radical extension $K$ of $F$ but that $[L:F]$ is not a power of $2$. Choose an odd prime $p$ dividing $[L:F]=G$, and choose an element $\tau \in G$ of order $p$. Then $\tau $ is not the identity^{}, so for some $i$, $\tau ({\alpha}_{i})\ne {\alpha}_{i}$. Also, since $f(x)$ is irreducible, $G$ acts transitively on the roots of $f(x)$, so for some $\nu \in G$, $\nu (\alpha )={\alpha}_{i}$. Then $\sigma ={\nu}^{1}\tau \nu $ does not fix $\alpha $, since
$${\nu}^{1}\tau \nu (\alpha )={\nu}^{1}\tau ({\alpha}_{i})\ne {\nu}^{1}({\alpha}_{i})=\alpha $$ 
Let $N={L}^{\sigma}$ be the fixed field of $\sigma $. Then $L$ is Galois over $N$, and clearly $[L:N]=p$. But Galois subfields of real radical extensions are at most quadratic, so $L$ cannot lie in a real radical extension of $N$.
However, $\alpha \notin N,\alpha \in L$, and $[L:N]$ is prime. Thus $L=N(\alpha )\subset NK$ (since $\alpha \in K$). Additionally, since $F\subset F(\alpha )\subset K$ is a real radical extension of $F$, we have also that $NK$ is a real radical extension of $NF=N$. So $L$ lies in the real radical extension $NK$ of $N$. But this is a contradiction^{} and thus $[L:F]$ must be a power of $2$.
One consequence of this theorem is the fact that if $f(x)\in F[x]$ has degree not a power of $2$, then if $f(x)$ has all real roots, those roots are not expressible in terms of real radicals. If $\mathrm{deg}f=3$, we recover the original casus irreducibilis.
References
 1 D.A. Cox, Galois Theory^{}, WileyInterscience, 2004.
Title  proof of casus irreducibilis for real fields 

Canonical name  ProofOfCasusIrreducibilisForRealFields 
Date of creation  20130322 17:43:08 
Last modified on  20130322 17:43:08 
Owner  rm50 (10146) 
Last modified by  rm50 (10146) 
Numerical id  8 
Author  rm50 (10146) 
Entry type  Theorem 
Classification  msc 12F10 