Del Ferro's depressed cubic equation solution

Del Ferro was the first person to have solved a variant of the cubic equation. This derivation is from Paul Nahin's book Imaginary Tale

He solved the equation $x^3+px=q$, where $p,q \in R,p,q >0$. Then he substituted for $x=m+n$

$ \Rightarrow (m+n)^3+p(m+n)=q$

$\Rightarrow m^3+n^3+3m^2n+3mn^2+pm+pn=q$

$\Rightarrow m^3+n^3+m(3mn+p)+n(3mn+p)=q$

$\Rightarrow m^3+n^3+(m+n)(3mn+p)=q$

Now is the magic step he puts $m^3+n^3=q, 3mn+p=0$

$\Rightarrow n= \frac{-p}{3m}$

$\Rightarrow m^3+\left (\frac{-p}{3m} \right)^3=q$

$\Rightarrow m^3-\frac{p^3}{27m^3}=q$

$\Rightarrow m^6-qm^3-\frac{p^3}{27m^3}=0$

As this is a quadratic equation in $ m^3$ by using the formula we get

$\Rightarrow m^3= \dfrac{q\pm \sqrt{q^2+\frac{4p^3}{27}}}{2}$

$\Rightarrow m^3= \dfrac{q\pm 2\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}{2}$

$\Rightarrow m^3= \frac{q}{2} \pm \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$

Taking positive first for $m^3$ we get

$\frac{q}{2} + \sqrt{\frac{q^2}{4}+\frac{p^3}{27}} +n^3=q$

$\Rightarrow n^3 =\frac{q}{2} - \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}$

Hence we have $m =\sqrt[3]{\frac{q}{2}+ \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$ and $n =\sqrt[3]{\frac{q}{2}- \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$

$\Rightarrow x=\sqrt[3]{\frac{q}{2}+ \sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$