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 $E=0$.
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3See this post. Even if several coefficients are zero, like $x^5-2x+1$, it cannot be solved by radicals in general. – Dietrich Burde Aug 06 '20 at 08:13
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4There is plenty of information in the Wikipedia article about quintic equations. I have collated this information in the tag wiki. – Toby Mak Aug 06 '20 at 08:14
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$E=0$ amounts to a quartic. $D=0$ is equivalent to $A=0$. – Aug 06 '20 at 08:41
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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 $B=C$.