# Through the Mind’s Eye

Let’s derive Cardano’s formula for the depressed cubic equation $x^3+px=q.$

Imaging a large cube with length $u.$ It is comprised of six blocks. Each block has its label on the top.

The large cube’s volume, $u^3$ is the sum of six smaller pieces:

$v^3+v^2(u-v) + v(u-v)u + v(u-v)u+(u-v)^2(u-v)+(u-v)^2v= u^3$

That is,

$(u-v)^3+\underline{2uv(u-v)}+\underline{v^2(u-v)}+\underline{(u-v)^2v}=u^3-v^3.$

Factoring $u-v$ from the underlined terms above gives

$(u-v)^3+(2uv + v^2 + (u-v)v)(u-v)=u^3-v^3$

or simply,

$(u-v)^3+\underbrace{3uv}_{p}(u-v)=\underbrace{u^3-v^3}_{q}.\quad\quad\quad(1)$

From the cube and (1), we see that for $u>v>0,$ if $3uv = p$ and $u^3-v^3=q,$ then $u-v,$ a positive quantity, is a solution of $x^3 + px =q.$

However, (1) is an algebraic identity regardless of the cube that led us to it. i.e.,

(1) is true for all $u, v \in R.$

Hence,

$\forall u,v \in R, (3uv = q, u^3-v^3=p) \implies u-v,$ not necessary a positive quantity, is a solution of depressed cubic $x^3+px=q.$

It follows that to find a solution of $x^3+px=q,$ we solve

$\begin{cases} 3uv=p \\ u^3-v^3=q \end{cases}$

for $u, v,$ after which the Cardano formula emerges:

$x=u-v=\sqrt[3]{\frac{q}{2}+\sqrt{\frac{q^2}{4} + \frac{p^3}{27}}} - \sqrt[3]{-\frac{q}{2} + \sqrt{\frac{q^2}{4} + \frac{p^3}{27}}}.\quad\quad\quad(\star)$

Exercise-1 Show that

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

after solving

$\begin{cases} 3uv=p \\ u^3-v^3=q\end{cases}$

for $u, v.$