FRACTURE

The quadratic served as practice. Now it’s time to see whether the same idea can survive in a far more hostile environment.

Just as before, the first objective is to eliminate a troublesome term and simplify the equation into a form that reveals its structure.

$\displaystyle ax^3 + bx^2 + cx + d = 0 \implies x^3 + \left(\frac{b}{a}\right)x^2 + \left(\frac{c}{a}\right)x + \left(\frac{d}{a}\right) = 0$

Here, $\ p = \dfrac{b}{a}, \quad q = \dfrac{c}{a}, \quad r = \dfrac{d}{a}$

Just as in the quadratic case, the goal is to shift the variable so that one entire term disappears.

$\displaystyle \text{Let, } \ x = t + k \implies (t+k)^3 + p(t+k)^2 + q(t+k) + r = 0$

$\phantom{\displaystyle \text{Let, } \ x = t + k} \implies t^3 + 3kt^2 + 3k^2t + k^3 + p(t^2 + 2kt + k^2) + qt + qk + r = 0$

$\phantom{\displaystyle \text{Let, } \ x = t + k} \implies t^3 + (3k + p)t^2 + (3k^2 + 2pk + q)t + (k^3 + pk^2 + qk + r) = 0$

To remove the $t^2$ term: $3k + p = 0 \implies k = -\dfrac{p}{3}$

$\displaystyle \therefore \ x = t - \frac{p}{3} \implies \left(t - \frac{p}{3}\right)^3 + p\left(t - \frac{p}{3}\right)^2 + q\left(t - \frac{p}{3}\right) + r = 0$

$\displaystyle \implies t^3 - 3t^2\cdot\frac{p}{3} + 3t\left(\frac{p}{3}\right)^2 - \left(\frac{p}{3}\right)^3 + \, pt^2 - 2tp\cdot\frac{p}{3} + p\cdot\frac{p^2}{9} + \, qt - q\cdot\frac{p}{3} + r = 0$

$\displaystyle \implies t^3 - \left(\frac{p^2}{3} - q\right)t + \left(\frac{p^3}{9} - \frac{p^3}{27} - \frac{pq}{3} + r\right) = 0$

$\displaystyle \implies t^3 + \left(-\frac{p^2}{3} + q\right)t + \left(\frac{2p^3}{27} - \frac{pq}{3} + r\right) = 0$

$\displaystyle \therefore \ \boxed{t^3 + p_1t + q_1 = 0}$ where, $\displaystyle p_1 = -\frac{p^2}{3} + q, \quad q_1 = \frac{2p^3}{27} - \frac{pq}{3} + r$

The quadratic lost its linear term, the cubic loses its quadratic term. This transformed equation is known as a depressed cubic.

But the equation is still not solved. The challenge now is to somehow untangle the remaining mixture of $t^3$ and $t$.

The next substitution looks almost impossible to guess. Unlike the Tschirnhaus shift, it doesn’t simplify by moving terms around—it simplifies through cancellation.

Let, $\displaystyle t = u + \frac{k_1}{u} \implies \left(u + \frac{k_1}{u}\right)^3 + p_1\left(u + \frac{k_1}{u}\right) + q_1 = 0$

$\phantom{\displaystyle \text{Let, } t = u + \frac{k_1}{u}} \implies u^3 + 3k_1 u + \dfrac{3k_1^2}{u} + \dfrac{k_1^3}{u^3} + p_1 u + \dfrac{p_1 k_1}{u} + q_1 = 0$

To eliminate the linear in $u$ term: $3k_1 u + p_1 u = 0 \implies k_1 = -\dfrac{p_1}{3}$

$\displaystyle \therefore \ t = u - \frac{p_1}{3u}$

$\displaystyle \implies \left(u - \frac{p_1}{3u}\right)^3 + p_1\left(u - \frac{p_1}{3u}\right) + q_1 = 0$

$\displaystyle \implies u^3 - 3u^2\left(\frac{p_1}{3u}\right) + 3u\left(\frac{p_1}{3u}\right)^2 - \left(\frac{p_1}{3u}\right)^3 + \, p_1u - \frac{p_1^2}{3u} + q_1 = 0$

$\displaystyle \implies u^3 - p_1u + \frac{p_1^2}{3u} - \frac{p_1^3}{27u^3} + p_1u - \frac{p_1^2}{3u} + q_1 = 0$

$\displaystyle \implies u^3 - \frac{p_1^3}{27u^3} + q_1 = 0$

$\displaystyle \implies u^6 - \frac{p_1^3}{27} + q_1u^3 = 0$

Surprisingly, the cubic has transformed into a quadratic in the quantity $u^3$.

$\displaystyle \implies (u^3)^2 + q_1(u^3) - \frac{p_1^3}{27} = 0$

$\displaystyle \implies u^3 = -\frac{q_1}{2} \pm \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}$ (Using the p-q formula)

$\displaystyle \therefore \ u = \sqrt[3]{-\frac{q_1}{2} \pm \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}}, \quad \omega\cdot\sqrt[3]{-\frac{q_1}{2} \pm \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}}, \quad \omega^2\cdot\sqrt[3]{-\frac{q_1}{2} \pm \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}}$

The remaining work is mostly algebra: substitute back, simplify carefully, and hope the structure holds together.

Although $u$ has a $\pm$ sign from the quadratic formula, plugging either the $+$ or $-$ version into $t = u - \dfrac{p_1}{3u}$ yields the exact same set of solutions.

$\displaystyle \bullet \quad \text{For } u = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} \ :$

$\displaystyle t = u + \frac{k_1}{u} = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \frac{-\dfrac{p_1}{3}}{\sqrt[3]{-\dfrac{q_1}{2} + \sqrt{\dfrac{q_1^2}{4} + \dfrac{p_1^3}{27}}}}$

$\displaystyle \implies t = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \frac{-\dfrac{p_1}{3} \left( \sqrt[3]{-\dfrac{q_1}{2} - \sqrt{\dfrac{q_1^2}{4} + \dfrac{p_1^3}{27}}} \right) }{\sqrt[3]{-\dfrac{q_1}{2} + \sqrt{\dfrac{q_1^2}{4} + \dfrac{p_1^3}{27}}} \left( \sqrt[3]{-\dfrac{q_1}{2} - \sqrt{\dfrac{q_1^2}{4} + \dfrac{p_1^3}{27}}} \right) }$

$\displaystyle \implies t = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \frac{-\dfrac{p_1}{3} \cdot \sqrt[3]{-\dfrac{q_1}{2} - \sqrt{\dfrac{q_1^2}{4} + \dfrac{p_1^3}{27}}}}{\sqrt[3]{-\dfrac{p_1^3}{27}}}$

$\displaystyle \implies t = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}}$

$\displaystyle \bullet \quad \text{Similarly for } u = \sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} \ :$

$\displaystyle t = \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}}$

Interestingly, the choice of $+$ or $-$ here for $u$ doesn’t produce a different value of $t$. The two cube-root terms simply trade places, leaving their sum unchanged.

Therefore, for a depressed cubic $\ t^3 + p_1t + q_1 = 0$

$$ \boxed{ \begin{aligned} t_1 &= \sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} \\[6pt] t_2 &= \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)\sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)\sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} \\[6pt] t_3 &= \left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right)\sqrt[3]{-\frac{q_1}{2} + \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} + \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)\sqrt[3]{-\frac{q_1}{2} - \sqrt{\frac{q_1^2}{4} + \frac{p_1^3}{27}}} \end{aligned} } $$

The remaining task is expressing every quantity in terms of the original coefficients.

Here, $\displaystyle p_1 = -\frac{p^2}{3} + q \qquad \left[ p = \frac{b}{a}, \quad q = \frac{c}{a} \right]$

$\displaystyle \implies p_1 = -\frac{(b/a)^2}{3} + \left(\frac{c}{a}\right) = -\frac{b^2}{3a^2} + \frac{c}{a}$

$\displaystyle \implies \frac{p_1^3}{27} = \frac{1}{27}\left(-\frac{b^2}{3a^2} + \frac{c}{a}\right)^3$

$\implies \boxed{\dfrac{p_1^3}{27} = \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}$

And, $\displaystyle q_1 = \frac{2p^3}{27} - \frac{pq}{3} + r \qquad \left[ p = \frac{b}{a}, \quad q = \frac{c}{a}, \quad r = \frac{d}{a}\right]$

$\displaystyle \implies q_1 = \frac{2(b/a)^3}{27} - \frac{(b/a)(c/a)}{3} + \left(\frac{d}{a}\right)$

$\displaystyle \implies q_1 = \frac{2b^3}{27a^3} - \frac{bc}{3a^2} + \frac{d}{a}$

$\displaystyle \implies \boxed{\frac{-q_1}{2} = \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}}$

$\implies \boxed{\frac{q_1^2}{4} = \left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2}$

The hard part is over. What remains is substitution back into the original coefficients.

$\displaystyle x = t + k = t - \frac{p}{3} = t - \frac{b}{3a} \qquad \left[ k = -\frac{p}{3}, \quad p = \frac{b}{a} \right]$

$\displaystyle \therefore \ \text{If } ax^3 + bx^2 + cx + d = 0,$

$$ \boxed{ \begin{aligned} x_1 = {} & \sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} + \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} \\ + &\sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} - \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} - \frac{b}{3a} \\[5pt] x_2 = {} & \left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right) \sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} + \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} \\ + &\left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right) \sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} - \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} - \frac{b}{3a} \\[5pt] x_3 = {} & \left(-\frac{1}{2} - \frac{i\sqrt{3}}{2}\right) \sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} + \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} \\ + &\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right) \sqrt[3]{\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} - \sqrt{\left(\frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}\right)^2 + \left(\frac{c}{3a} -\frac{b^2}{9a^2}\right)^3}} - \frac{b}{3a}\end{aligned} } $$

The final formula is enormous, but the path leading to it is surprisingly simple in hindsight.

First remove the quadratic term. Then transform the depressed cubic into a quadratic hidden inside $u^3$. Solve that quadratic, and unwind the substitutions.

What began as a seemingly impossible formula turned out to be a sequence of familiar ideas wearing increasingly elaborate disguises—and that curiosity now points me toward the quartic formula—eager to see if it gives up its secrets just as readily.