## An Icosahedral and Brioschi quintic identity

Here’s an identity I found.  For arbitrary r, define,

$a = \frac{r^5(r^{10}+11r^5-1)^5}{(r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1)^2}$

and,

$w = \frac{r^2(r^{10}+11r^5-1)^2(r^6+2r^5-5r^4-5r^2-2r+1)}{r^{30}+522r^{25}-10005r^{20}-10005r^{10}-522r^5+1}$

then,

$w^5-10aw^3+45a^2w-a^2 = 0$

Those two complicated expressions neatly wrap up into that last equation, doesn’t it?  This is the Brioschi quintic form which the general quintic can be reduced into.  Two of the polynomials are easily recognizable as icosahedral invariants, while,

$P(r) = r^6+2r^5-5r^4-5r^2-2r+1$

is a polynomial invariant for the octahedron. This gave rise to the question here.