Root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$ What is real root of the quintic $x^5 − 5x^4 + 30x^3 − 50x^2 + 55x − 21=0$?
Some remarks:


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*I saw this quintic in wikipedia

*Real root is given $x=1+{\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$ in wikipedia. 

*I used the transformation $x=y+1$ (Tschirnhaus transformation) and $y^5 + 20 y^3 + 20 y^2 + 30 y + 10=0$. (We can remove the term of degree four.)

*Therefore, we have to solve $x^5 + 20 x^3 + 20 x^2 + 30 x + 10=0$ and we have to find $x={\sqrt[ {5}]{2}}-\left({\sqrt[ {5}]{2}}\right)^{2}+\left({\sqrt[ {5}]{2}}\right)^{3}-\left({\sqrt[ {5}]{2}}\right)^{4}$. 

*But, I want to know how to solve this without plugging it in and verifying an already known root. Can the depressed quintic be solved? Or does one need to use another method to solve this polynomial?

First off, I don't know of a systematic way to solve it. But suppose one has the magic inspiration to effect the substitution $x=y+2\,$, then the equation in $y$ turns out to be: $$y^5 + 5 y^4 + 30 y^3 + 90 y^2 + 135 y + 81 = 0$$
At this point, the equation became easy to solve. The ratios between symmetric coefficients are $90/30=3^1\,$, $135/5=3^3\,$, $81/1=3^4\,$, suggesting the substitution $y=3z\,$, which gives: $$3 z^5 + 5 z^4 + 10 z^3 + 10 z^2 + 5 z + 1=0 \;\;\iff\;\; 2z^5 + (z+1)^5=0$$
The latter has the obvious real solution $1+ \cfrac{1}{z}=-\sqrt[5]{2}\,$ which can be rationalized as: $$z = \frac{-1}{1+\sqrt[5]{2}} \cdot \frac{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4}{1-\sqrt[5]{2}+\sqrt[5]{2}^2-\sqrt[5]{2}^3+\sqrt[5]{2}^4} = \frac{-1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4}{1+2} $$
Reverting back to $x=3z+2$ gives $x=1+\sqrt[5]{2}-\sqrt[5]{2}^2+\sqrt[5]{2}^3-\sqrt[5]{2}^4\,$.
