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The quintic can be transformed to the one-parameter Brioschi quintic,

$$u^5-10\alpha u^3+45\alpha^2u-\alpha^2 = 0\tag1$$

This form is well-known for its connection to the symmetries of the icosahedron. I found that using a similar transformation, the general quintic can be reduced, also in radicals, to,

$$v^5-5\beta v^3+10\beta^2v-\beta^2 = 0\tag2$$

Question: Does $(2)$ appear anywhere when studying icosahedral symmetry or similar objects? If not, then what is the reason why the quintic can be reduced, in radicals, to this one-parameter form?

P.S. Incidentally, doing the transformation $u = 1/(x^2+20)$ on $(1)$ and $v = 4/(y^2+15)$ on $(2)$ reduces them to the rather nice similar forms,

$$(x^2+20)^2(x-5)+\frac{1}{\alpha}=0\tag3$$

$$(y^2+15)^2(y-5)+\frac{32}{\beta}=0\tag4$$

Tito Piezas III
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Geometry behind the Brioschi quintic is explained in Jerry Shurman book "Geometry of the Quintic" (available from his website http://people.reed.edu/~jerry/ ). However, it seems no explanation of (2) is given there.

Zurab Silagadze
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    My feeling is that since it is not trivial to reduce the quintic to one-parameter form, then there *has* to be some reason (geometric or other) why it can transformed into something so similar to the Brioschi form. Note that it is also very reminiscent to the solvable DeMoivre quintic $$x^5-5\alpha, x^3+5\alpha^2 x+\beta = 0$$ – Tito Piezas III Dec 07 '15 at 06:34