Recently I found this lovely algebraic equation solver flowchart online. What I do not understand is how to get from the Depressed Quartic (top right) to the Monic Cubic polynomial (the path where $\alpha_1 \neq 0$). I see that the intermediate equation $(u^2+[\alpha_2+y])^2=\dots$ is basically equivalent to the depressed quartic equation, but the next step is a mystery. What exactly do I have to do there? Is the $y$ in this equation the same as the one in th emonic cubic polynomial?
For arbitrary $y$, you have
$$ (u^2+[\alpha_2+y])^2= \delta_2 u^2 + \delta_1 u + \delta_0 $$ with $$ \delta_2 = \alpha_2 + 2y, \delta_1 = -\alpha_1, \delta_0 = [\alpha_2+y]^2 - \alpha_0 \quad . $$ Now $y$ is determined such that $\delta_1^2 - 4 \delta_2 \delta_0 = 0$, this gives a cubic equation for $y$.
If you substitute this solution $y$ back into the first equation, the right-hand side becomes a "perfect square"
$$ (u^2+[\alpha_2+y])^2= \frac 1{4\delta_0}(\delta_1 u + 2\delta_0) ^2 $$ and you can proceed by taking square roots: $$ u^2+[\alpha_2+y] = \pm \frac 1{2\sqrt{\delta_0}}(\delta_1 u + 2\delta_0) $$