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.