Solutions to quintic equation

Question

Abel's Theorem shows that there is no general formula that gives the solution to $x^5+Ax^4+Bx^3+Cx^2+Dx+E=0$ with radicals. My math is not sufficient to understand Abel's theorem. Considering motivation from lower degrees where we try to remove terms, e.g. depressing a cubic, does it(or other theorems) say anything about the existence of special case formulas in radicals? For example, $D=0$ or $B=C$. I am less interested in the case $E=0$.

Answer 1

By shifting the unknown by $-\dfrac A4$, you deplete the equation and absorb one degree of freedom. So introducing a constraint such as $D=0$ or $B=C$ does not make the equation less general.

$D=0$ is what you get by changing $x\leftrightarrow \dfrac1x$ and depleting.

By resolution of a cubic, you can find a shift the enforces $B=C$.