casus irreducibilis

Let the polynomial

$$P(x):=x^n+a_1{x}^{n-1}+\mathrm{\dots }+a_n$$

with complex coefficients $a_j$ be irreducible (http://planetmath.org/IrreduciblePolynomial2), i.e. irreducible in the field $\mathbb{Q}(a_1,\mathrm{\dots },a_n)$  of its coefficients.  If the equation  $P(x)=0$  can be solved algebraically (http://planetmath.org/AlgebraicallySolvable) and if all of its roots are real, then no root may be expressed with the numbers $a_j$ using mere real radicals (http://planetmath.org/NthRoot) unless the degree (http://planetmath.org/AlgebraicEquation) $n$ of the equation is an integer power (http://planetmath.org/GeneralAssociativity) of 2.