MHB Solution to function with power greater than 4

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Why we can't use radical to solve an equations with power greater than 4?
 
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You can solve certain equations with higher than 4:

  1. $x^n-a=0$
  2. $x^{2n}+x^n+a=0$
  3. $x^{3n}+x^{2n}+x^n+a=0$
  4. $x^{4n}+x^{3n}+x^{2n}+x^n+a=0$
  5. etc.
where $a \in \Bbb{R}$ and $n\in \Bbb{N}$.

However, the general fifth degree or higher polynomial does not have a radical solution due to a theorem, Abel-Ruffin Theorem. there is no solutions in radical to general polynomial equations of degree five or higher with arbitrary coefficients.

Note: it does not assert some higher-degree polynomials have no solutions... (see at the beginning of the post)
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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