<s>[INST] How do I find if $\frac{e^x}{x^3} = 2x + 1$ has an algebraic solution? Is there some way of solving $$\frac{e^x}{x^3} = 2x + 1 $$
non-numerically?
How would I go about proving if there exists a closed form solution? Similarly how would I go about proving if there exists an analytic solution?
 [/INST] If you write your equation as, say, $f(x)=t$ where $f(x) = e^x - 2 x^4 - x^3$, you can use the Lagrange inversion formula to get power series solutions.  A convenient starting point is $f(1)=e-3$.  Taking $a=1$ and $b=f(a) = e-3$, the Lagrange inversion formula says that for $t$ near $b$, one solution of $f(x)=t$ is 
$$ g(t) = a + \frac{t-b}{f'(a)} + \sum_{n=2}^\infty \lim_{w \to a} \frac{d^{n-1}}{dw^{n-1}}
\left(\frac{w-a}{f(w)-b}\right)^n \frac{(t-b)^n}{n!}$$ 
which in this case is
$$ g(t) = 1 + \frac{t-b}{e-11} + \frac{-e+30}{(e-11)^3} \frac{(t-b)^2}{2} + \frac{2 e^2 - 115 e +2106}{(e-11)^5} \frac{(t-b)^3}{6} + \frac{-6e^3+470 e^2 - 16237 e + 232608}{(e-11)^7} \frac{(t-b)^4}{24} + \ldots$$
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