$x^5-6x+3$ over $\mathbb{Q}$ is not solvable by radicals

The prime $3$ divides the coefficients $-6$ and $3$, does not divide the leading coefficient $1$, and $3^2=9$ does not divide the constant term $3$. So [L1] makes $f$ irreducible over $\mathbb{Q}$.

Let $a:=(6\mathrm{/}5{)}^{1\mathrm{/}4}>1$, which exists by [L2]. For real numbers $x<y$ one has $f(y)-f(x)=(y-x)(y^4+y^{3}x+y^2{x}^2+y{x}^3+x^4-6)\mathrm{.}$ If $a\le x<y$ or $x<y\le -a$, then each of the five degree-four monomials in parentheses is at least $a^4=6\mathrm{/}5$, and at least one is strictly larger than $6\mathrm{/}5$; hence the parenthesis is $>5a^4-6=0$, so $f(y)>f(x)$. If $-a\le x<y\le a$, then each of those monomials is at most $a^4=6\mathrm{/}5$, and they cannot all equal $6\mathrm{/}5$ when $x<y$: equality in the $x^4$ and $y^4$ terms would force $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }=a$, while $x<y$ would then give $(x,y)=(-a,a)$ and hence $y^{3}x=-a^4$. So the parenthesis is $<5a^4-6=0$, and therefore $f(y)<f(x)$. Thus $f$ is increasing on $(-\mathrm{\infty },-a]$, decreasing on $[-a,a]$, and increasing on $[a,\mathrm{\infty })$.

Because $a^4=6\mathrm{/}5<16=2^4$ and $a>0$, one has $a<2$. Also $f(-a)=\frac{24a}{5}+3>0\text{and}f(a)=3-\frac{24a}{5}<0.$ Together with $f(-2)=-17<0$ and $f(2)=23>0$, the continuity from [L3] and the intermediate value theorem [L5] give a root in each of the three intervals $(-2,-a),(-a,a),(a,2)\mathrm{.}$ Step 1.2 shows that $f$ is monotone on each of the three corresponding regions, so there is at most one root in each. Therefore $f$ has exactly three real roots.

By [L6], choose all five roots of $f$ in $\mathbb{C}$ and let $E\subseteq \mathbb{C}$ be the subfield they generate over $\mathbb{Q}$. Fact [L11] makes $E$ a splitting field of $f$ over $\mathbb{Q}$. Step 2.1 shows that exactly three of those roots are real, so the remaining two roots are nonreal. Complex conjugation on $\mathbb{C}$ fixes $\mathbb{Q}$ and preserves the root set, hence it restricts to a $\mathbb{Q}$-automorphism of $E$ that fixes the three real roots and swaps the two nonreal roots. Thus $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ contains a transposition.

The field $\mathbb{Q}$ has characteristic $0$, so [L4] makes it perfect. Step 1.1 shows that $f$ is irreducible over $\mathbb{Q}$, and therefore $f$ is separable by the defining property of a perfect field in [L4]. Fact [L7] now makes $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ transitive on the five roots. With step 3.1, fact [L8] gives $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})=S_5\mathrm{.}$

By [L9], the group $S_5$ is not solvable. If $f$ were solvable by radicals, [L10] would force $\mathrm{Gal}(E\mathrm{/}\mathbb{Q})$ to be solvable, contradicting step 4.1. Hence $f$ is not solvable by radicals.